Difference between revisions of "The Limit Laws"

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Now that we have the formal definition of a limit, we can set about proving some of the properties we stated earlier in this chapter about limits.
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===Constant Rule for Limits===
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: If <math>a,b</math> are constants then <math>\lim_{x\to a}b=b</math>.
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Proof of the Constant Rule for Limits: We need to find a <math>\delta>0</math> such that for every <math>\varepsilon>0</math> , <math>|b-b|<\varepsilon</math> whenever <math>0<|x-a|<\delta</math> . <math>|b-b|=0</math> and <math>\varepsilon>0</math> , so <math>|b-b|<\varepsilon</math> is satisfied independent of any value of <math>\delta</math> ; that is, we can choose any <math>\delta</math> we like and the <math>\varepsilon</math> condition holds.
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===Identity Rule for Limits===
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: If <math>a</math> is a constant then <math>\lim_{x\to a}x=a</math>.
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Proof: To prove that <math>\lim_{x\to a}x=a</math> , we need to find a <math>\delta>0</math> such that for every <math>\varepsilon>0</math> , <math>|x-a|<\varepsilon</math> whenever <math>0<|x-a|<\delta</math> . Choosing <math>\delta=\varepsilon</math> satisfies this condition.
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===Scalar Product Rule for Limits===
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: Suppose that <math>\lim_{x\to a}f(x)=L</math> for finite <math>L</math> and that <math>c</math> is constant. Then <math>\lim_{x\to a}c\cdot f(x)=c\cdot\lim_{x\to a}f(x)=c\cdot L</math>}}.
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Proof: Given the limit above, there exists in particular a <math>\delta>0</math> such that <math>\Big|f(x)-L\Big|<\frac{\varepsilon}{k}</math> whenever <math>0<|x-a|<\delta</math> , for some <math>k>0</math> such that <math>|c|<k</math> . Hence
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:<math>\Big|c\cdot f(x)-c\cdot L\Big|=|c|\cdot\Big|f(x)-L\Big|<k\cdot\frac{\varepsilon}{k}=\varepsilon</math>.
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===Sum Rule for Limits===
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: Suppose that <math>\lim_{x\to c}f(x)=L</math> and <math>\lim_{x\to c}g(x)=M</math>. Then,
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:: <math>\lim_{x\to c}\Big[f(x)+g(x)\Big]=\lim_{x\to c}f(x)+\lim_{x\to c}g(x)=L+M</math>}}.
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Proof: Since we are given that <math>\lim_{x\to c}f(x)=L</math> and <math>\lim_{x\to c}g(x)=M</math> , there must be functions, call them <math>\delta_f(\varepsilon)</math> and <math>\delta_g(\varepsilon)</math> , such that for all <math>\varepsilon>0</math> , <math>\Big|f(x)-L\Big|<\varepsilon</math> whenever <math>|x-c|<\delta_f(\varepsilon)</math> , and <math>\Big|g(x)-M\Big|<\varepsilon</math> whenever <math>|x-c|<\delta_{g}(\varepsilon)</math> .<br/>
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Adding the two inequalities gives <math>\Big|f(x)-L\Big|+\big|g(x)-M\big|<2\varepsilon</math> . By the triangle inequality we have <math>\bigg|(f(x)-L)+(g(x)-M)\bigg|=\bigg|(f(x)+g(x))-(L+M)\bigg|\le\Big|f(x)-L\Big|+\Big|g(x)-M\Big|</math> , so we have <math>\bigg|(f(x)+g(x))-(L+M)\bigg|<2\varepsilon</math> whenever <math>|x-c|<\delta_f(\varepsilon)</math> and <math>|x-c|<\delta_{g}(\varepsilon)</math> . Let <math>\delta_{fg}(\varepsilon)</math> be the smaller of <math>\delta_f(\tfrac{\varepsilon}{2})</math> and <math>\delta_g(\tfrac{\varepsilon}{2})</math> . Then this <math>\delta</math> satisfies the definition of a limit for <math>\lim_{x\to c}\Big[f(x)+g(x)\Big]</math> having limit <math>L+M</math> .
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===Difference Rule for Limits===
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: Suppose that <math>\lim_{x\to c}f(x)=L</math> and <math>\lim_{x\to c}g(x)=M</math>. Then
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:: <math>\lim_{x\to c}\Big[f(x)-g(x)\Big]=\lim_{x\to c}f(x)-\lim_{x\to c}g(x)=L-M</math>}}
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Proof: Define <math>h(x)=-g(x)</math> . By the Scalar Product Rule for Limits, <math>\lim_{x\to c}h(x)=-M</math> . Then by the Sum Rule for Limits, <math>\lim_{x\to c}\Big[f(x)-g(x)\Big]=\lim_{x\to c}\Big[f(x)+h(x)\Big]=L-M</math>.
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===Product Rule for Limits===
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: Suppose that <math>\lim_{x\to c}f(x)=L</math> and <math>\lim_{x\to c}g(x)=M</math> . Then
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:: <math>\lim_{x\to c}\Big[f(x)\cdot g(x)\Big]=\lim_{x\to c}f(x)\cdot\lim_{x\to c}g(x)=L\cdot M</math>}}
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Proof: Let <math>\varepsilon</math> be any positive number. The assumptions imply the existence of the positive numbers <math>\delta_1,\delta_2,\delta_3</math> such that<br/>
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:<math>(1)\qquad\Big|f(x)-L\Big|<\frac{\varepsilon}{2(1+|M|)}</math> when <math>0<|x-c|<\delta_1</math>
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:<math>(2)\qquad\Big|g(x)-M\Big|<\frac{\varepsilon}{2(1+|L|)}</math> when <math>0<|x-c|<\delta_2</math>
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:<math>(3)\qquad\Big|g(x)-M\Big|<1</math> when <math>0<|x-c|<\delta_3</math>
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According to the condition (3) we see that
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:<math>\Big|g(x)\Big|=\bigg|g(x)-M+M\bigg|\le\Big|g(x)-M\Big|+|M|<1+|M|</math> when <math>0<|x-c|<\delta_3</math>
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Supposing then that <math>0<|x-c|<\min\{\delta_1,\delta_2,\delta_3\}</math> and using (1) and (2) we obtain
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:<math>\begin{align}\bigg|f(x)g(x)-LM\bigg|
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&=\bigg|f(x)g(x)-Lg(x)+Lg(x)-LM\bigg|\\
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&\le\bigg|f(x)g(x)-Lg(x)\bigg|+\bigg|Lg(x)-LM\bigg|\\
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&=\Big|g(x)\Big|\cdot\Big|f(x)-L\Big|+|L|\cdot\Big|g(x)-M\Big|\\
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&<(1+|M|)\frac{\varepsilon}{2(1+|M|)}+(1+|L|)\frac{\varepsilon}{2(1+|L|)}\\
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&=\varepsilon
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\end{align}</math>
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===Quotient Rule for Limits===
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: Suppose that <math>\lim_{x\to c}f(x)=L</math> and <math>\lim_{x\to c}g(x)=M</math> and <math>M\ne 0</math>. Then
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:: <math>\lim_{x\to c}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x\to c}f(x)}{\lim\limits_{x\to c}g(x)}=\frac{L}{M}</math>
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Proof: If we can show that <math>\lim_{x\to c}\frac{1}{g(x)}=\frac{1}{M}</math> , then we can define a function, <math>h(x)</math> as <math>h(x)=\frac{1}{g(x)}</math> and appeal to the Product Rule for Limits to prove the theorem. So we just need to prove that <math>\lim_{x\to c}\frac{1}{g(x)}=\frac{1}{M}</math>.
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Let <math>\varepsilon</math> be any positive number. The assumptions imply the existence of the positive numbers <math>\delta_1,\delta_2</math> such that<br/>
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:<math>(1)\qquad\Big|g(x)-M\Big|<\frac{\varepsilon|M|^2}{2}</math> when <math>0<|x-c|<\delta_1</math>
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:<math>(2)\qquad\Big|g(x)-M\Big|<\frac{|M|}{2}</math> when <math>0<|x-c|<\delta_{2}</math>
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According to the condition (2) we see that
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:<math>|M|=|M-g(x)+g(x)|\le|M-g(x)|+|g(x)|<\frac{|M|}{2}+|g(x)| </math> so <math> |g(x)|>\frac{|M|}{2} </math> when <math>0<|x-c|<\delta_2</math><br/>
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which implies that
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:<math>(3)\qquad\left|\frac{1}{g(x)}\right|<\frac{2}{|M|}</math> when <math>0<|x-c|<\delta_2</math><br/>
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Supposing then that <math>0<|x-c|<\min\{\delta_1,\delta_2\}</math> and using (1) and (3) we obtain
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:<math>\begin{align}\left|\frac{1}{g(x)}-\frac{1}{M}\right|&=\left|\frac{M-g(x)}{Mg(x)}\right|\\
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&=\left|\frac{g(x)-M}{Mg(x)}\right|\\
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&=\left|\frac{1}{g(x)}\right|\cdot\left|\frac{1}{M}\right|\cdot|g(x)-M|\\
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&<\frac{2}{|M|}\cdot\frac{1}{|M|}\cdot|g(x)-M|\\
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&<\frac{2}{|M|}\cdot\frac{1}{|M|}\cdot\frac{\varepsilon|M|^2}{2}\\
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&=\varepsilon
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\end{align}</math>
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===Squeeze Theorem===
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: Suppose that <math>g(x)\le f(x)\le h(x)</math> holds for all <math>x</math> in some open interval containing <math>c</math> , except possibly at <math>x=c</math> itself. Suppose also that <math>\lim_{x\to c}g(x)=\lim_{x\to c}h(x)=L</math> . Then <math>\lim_{x\to c}f(x)=L</math> also.
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Proof: From the assumptions, we know that there exists a <math>\delta</math> such that <math>\Big|g(x)-L\Big|<\varepsilon</math> and <math>\Big|h(x)-L\Big|<\varepsilon</math> when <math>0<|x-c|<\delta</math> .<br/>
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These inequalities are equivalent to <math>L-\varepsilon<g(x)<L+\varepsilon</math> and <math>L-\varepsilon<h(x)<L+\varepsilon</math> when <math>0<|x-c|<\delta</math>.<br/>
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Using what we know about the relative ordering of <math>f(x),g(x)</math> , and <math>h(x)</math> , we have<br/>
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: <math>L-\varepsilon<g(x)\le f(x)\le h(x)<L+\varepsilon</math> when <math>0<|x-c|<\delta</math> .<br/>
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Then<br/>
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: <math>-\varepsilon<f(x)-L<\varepsilon</math> when <math>0<|x-c|<\delta</math> .<br/>
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So<br/>
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: <math>\Big|f(x)-L\Big|<\varepsilon</math> when <math>0<|x-c|<\delta</math> .
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==Resources==
 
* [https://mathresearch.utsa.edu/wikiFiles/MAT1214/The%20Limit%20Laws/MAT1214-2.3TheLimitLawsPwPt.pptx  The Limit Laws] PowerPoint file created by Dr. Sara Shirinkam, UTSA.
 
* [https://mathresearch.utsa.edu/wikiFiles/MAT1214/The%20Limit%20Laws/MAT1214-2.3TheLimitLawsPwPt.pptx  The Limit Laws] PowerPoint file created by Dr. Sara Shirinkam, UTSA.
  
 
* [https://mathresearch.utsa.edu/wikiFiles/MAT1214/The%20Limit%20Laws/MAT1214_2.3TheLimitLawsWS1.pdf  The Limit Laws Worksheet]
 
* [https://mathresearch.utsa.edu/wikiFiles/MAT1214/The%20Limit%20Laws/MAT1214_2.3TheLimitLawsWS1.pdf  The Limit Laws Worksheet]
−
 
  
 
* [https://youtu.be/P7G7F3NzPw0 Properties of Limits] by The Organic Chemistry Tutor
 
* [https://youtu.be/P7G7F3NzPw0 Properties of Limits] by The Organic Chemistry Tutor

Revision as of 15:45, 28 September 2021

Now that we have the formal definition of a limit, we can set about proving some of the properties we stated earlier in this chapter about limits.

Constant Rule for Limits

If are constants then .

Proof of the Constant Rule for Limits: We need to find a Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta>0} such that for every Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varepsilon>0} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |b-b|<\varepsilon} whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-a|<\delta} . Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |b-b|=0} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varepsilon>0} , so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |b-b|<\varepsilon} is satisfied independent of any value of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta}  ; that is, we can choose any Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta} we like and the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varepsilon} condition holds.

Identity Rule for Limits

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} is a constant then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to a}x=a} .

Proof: To prove that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to a}x=a} , we need to find a Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta>0} such that for every Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varepsilon>0} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-a|<\varepsilon} whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-a|<\delta} . Choosing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=\varepsilon} satisfies this condition.

Scalar Product Rule for Limits

Suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to a}f(x)=L} for finite Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} and that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} is constant. Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to a}c\cdot f(x)=c\cdot\lim_{x\to a}f(x)=c\cdot L} }}.

Proof: Given the limit above, there exists in particular a Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta>0} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Big|f(x)-L\Big|<\frac{\varepsilon}{k}} whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-a|<\delta} , for some Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k>0} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |c|<k} . Hence

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Big|c\cdot f(x)-c\cdot L\Big|=|c|\cdot\Big|f(x)-L\Big|<k\cdot\frac{\varepsilon}{k}=\varepsilon} .

Sum Rule for Limits

Suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}f(x)=L} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}g(x)=M} . Then,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}\Big[f(x)+g(x)\Big]=\lim_{x\to c}f(x)+\lim_{x\to c}g(x)=L+M} }}.

Proof: Since we are given that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}f(x)=L} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}g(x)=M} , there must be functions, call them Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_f(\varepsilon)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_g(\varepsilon)} , such that for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varepsilon>0} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Big|f(x)-L\Big|<\varepsilon} whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-c|<\delta_f(\varepsilon)} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Big|g(x)-M\Big|<\varepsilon} whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-c|<\delta_{g}(\varepsilon)} .
Adding the two inequalities gives Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Big|f(x)-L\Big|+\big|g(x)-M\big|<2\varepsilon} . By the triangle inequality we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bigg|(f(x)-L)+(g(x)-M)\bigg|=\bigg|(f(x)+g(x))-(L+M)\bigg|\le\Big|f(x)-L\Big|+\Big|g(x)-M\Big|} , so we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bigg|(f(x)+g(x))-(L+M)\bigg|<2\varepsilon} whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-c|<\delta_f(\varepsilon)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-c|<\delta_{g}(\varepsilon)} . Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_{fg}(\varepsilon)} be the smaller of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_f(\tfrac{\varepsilon}{2})} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_g(\tfrac{\varepsilon}{2})} . Then this satisfies the definition of a limit for having limit .

Difference Rule for Limits

Suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}f(x)=L} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}g(x)=M} . Then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}\Big[f(x)-g(x)\Big]=\lim_{x\to c}f(x)-\lim_{x\to c}g(x)=L-M} }}

Proof: Define Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)=-g(x)} . By the Scalar Product Rule for Limits, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}h(x)=-M} . Then by the Sum Rule for Limits, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}\Big[f(x)-g(x)\Big]=\lim_{x\to c}\Big[f(x)+h(x)\Big]=L-M} .

Product Rule for Limits

Suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}f(x)=L} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}g(x)=M} . Then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}\Big[f(x)\cdot g(x)\Big]=\lim_{x\to c}f(x)\cdot\lim_{x\to c}g(x)=L\cdot M} }}

Proof: Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varepsilon} be any positive number. The assumptions imply the existence of the positive numbers Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_1,\delta_2,\delta_3} such that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1)\qquad\Big|f(x)-L\Big|<\frac{\varepsilon}{2(1+|M|)}} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|<\delta_1}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (2)\qquad\Big|g(x)-M\Big|<\frac{\varepsilon}{2(1+|L|)}} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|<\delta_2}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (3)\qquad\Big|g(x)-M\Big|<1} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|<\delta_3}

According to the condition (3) we see that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Big|g(x)\Big|=\bigg|g(x)-M+M\bigg|\le\Big|g(x)-M\Big|+|M|<1+|M|} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|<\delta_3}

Supposing then that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|<\min\{\delta_1,\delta_2,\delta_3\}} and using (1) and (2) we obtain

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align}\bigg|f(x)g(x)-LM\bigg| &=\bigg|f(x)g(x)-Lg(x)+Lg(x)-LM\bigg|\\ &\le\bigg|f(x)g(x)-Lg(x)\bigg|+\bigg|Lg(x)-LM\bigg|\\ &=\Big|g(x)\Big|\cdot\Big|f(x)-L\Big|+|L|\cdot\Big|g(x)-M\Big|\\ &<(1+|M|)\frac{\varepsilon}{2(1+|M|)}+(1+|L|)\frac{\varepsilon}{2(1+|L|)}\\ &=\varepsilon \end{align}}

Quotient Rule for Limits

Suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}f(x)=L} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}g(x)=M} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M\ne 0} . Then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x\to c}f(x)}{\lim\limits_{x\to c}g(x)}=\frac{L}{M}}

Proof: If we can show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}\frac{1}{g(x)}=\frac{1}{M}} , then we can define a function, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)} as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)=\frac{1}{g(x)}} and appeal to the Product Rule for Limits to prove the theorem. So we just need to prove that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}\frac{1}{g(x)}=\frac{1}{M}} .

Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varepsilon} be any positive number. The assumptions imply the existence of the positive numbers Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_1,\delta_2} such that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1)\qquad\Big|g(x)-M\Big|<\frac{\varepsilon|M|^2}{2}} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|<\delta_1}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (2)\qquad\Big|g(x)-M\Big|<\frac{|M|}{2}} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|<\delta_{2}}

According to the condition (2) we see that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |M|=|M-g(x)+g(x)|\le|M-g(x)|+|g(x)|<\frac{|M|}{2}+|g(x)| } so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |g(x)|>\frac{|M|}{2} } when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|<\delta_2}

which implies that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (3)\qquad\left|\frac{1}{g(x)}\right|<\frac{2}{|M|}} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|<\delta_2}

Supposing then that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|<\min\{\delta_1,\delta_2\}} and using (1) and (3) we obtain

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align}\left|\frac{1}{g(x)}-\frac{1}{M}\right|&=\left|\frac{M-g(x)}{Mg(x)}\right|\\ &=\left|\frac{g(x)-M}{Mg(x)}\right|\\ &=\left|\frac{1}{g(x)}\right|\cdot\left|\frac{1}{M}\right|\cdot|g(x)-M|\\ &<\frac{2}{|M|}\cdot\frac{1}{|M|}\cdot|g(x)-M|\\ &<\frac{2}{|M|}\cdot\frac{1}{|M|}\cdot\frac{\varepsilon|M|^2}{2}\\ &=\varepsilon \end{align}}


Squeeze Theorem

Suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)\le f(x)\le h(x)} holds for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} in some open interval containing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} , except possibly at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=c} itself. Suppose also that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}g(x)=\lim_{x\to c}h(x)=L} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to c}f(x)=L} also.

Proof: From the assumptions, we know that there exists a such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Big|g(x)-L\Big|<\varepsilon} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Big|h(x)-L\Big|<\varepsilon} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|<\delta} .
These inequalities are equivalent to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L-\varepsilon<g(x)<L+\varepsilon} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L-\varepsilon<h(x)<L+\varepsilon} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|<\delta} .
Using what we know about the relative ordering of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x),g(x)} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)} , we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L-\varepsilon<g(x)\le f(x)\le h(x)<L+\varepsilon} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|<\delta} .

Then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\varepsilon<f(x)-L<\varepsilon} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|<\delta} .

So

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Big|f(x)-L\Big|<\varepsilon} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|<\delta} .


Resources

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