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]
  
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* [https://youtu.be/P7G7F3NzPw0 Properties of Limits] by The Organic Chemistry Tutor
  
* [https://youtu.be/mQpU2nYG3Jw The Limit Laws Part 1] Video Lecture by Instructor Sharon, UTSA.
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* [https://youtu.be/v_Nz6UUQ4HQ Limit Laws to Evaluate a Limit , Example 1] by patrickJMT
 
 
* [https://youtu.be/hquujjjX-3A The Limit Laws Part 2] Video Lecture by Instructor Sharon, UTSA.
 
  
* [https://youtu.be/DvND8hoNtHE The Limit Laws Part 3] Video Lecture by Instructor Sharon, UTSA.
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* [https://youtu.be/3d7qLvgZp0s Limit Laws to Evaluate a Limit , Example 2] by patrickJMT
  
* [https://youtu.be/KoxjCjrTrik The Limit Laws Part 4] Video Lecture by Instructor Sharon, UTSA.
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* [https://youtu.be/KeAgfb3NZLc Limit Laws to Evaluate a Limit , Example 3] by patrickJMT
  
* [https://youtu.be/jJsQxqU50Lw The Limit Laws Part 5] Video Lecture by Instructor Sharon, UTSA.
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==Licensing==
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Content obtained and/or adapted from:
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* [https://en.wikibooks.org/wiki/Calculus/Proofs_of_Some_Basic_Limit_Rules Proofs of Some Basic Limit Rules, Wikibooks: Calculus] under a CC BY-SA license

Latest revision as of 17:29, 15 January 2022

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 such that for every , whenever . and , so is satisfied independent of any value of  ; that is, we can choose any we like and the condition holds.

Identity Rule for Limits

If is a constant then .

Proof: To prove that , we need to find a such that for every , whenever . Choosing satisfies this condition.

Scalar Product Rule for Limits

Suppose that for finite and that is constant. Then .

Proof: Given the limit above, there exists in particular a such that whenever , for some such that . Hence

.

Sum Rule for Limits

Suppose that and . Then,
.

Proof: Since we are given that and , there must be functions, call them and , such that for all , whenever , and whenever .
Adding the two inequalities gives . By the triangle inequality we have , so we have whenever and . Let be the smaller of and . Then this satisfies the definition of a limit for having limit .

Difference Rule for Limits

Suppose that and . Then

Proof: Define . By the Scalar Product Rule for Limits, . Then by the Sum Rule for Limits, .

Product Rule for Limits

Suppose that and . Then

Proof: Let be any positive number. The assumptions imply the existence of the positive numbers such that

when
when
when

According to the condition (3) we see that

when

Supposing then that and using (1) and (2) we obtain

Quotient Rule for Limits

Suppose that and and . Then

Proof: If we can show that , then we can define a function, as and appeal to the Product Rule for Limits to prove the theorem. So we just need to prove that .

Let be any positive number. The assumptions imply the existence of the positive numbers such that

when
when

According to the condition (2) we see that

so when

which implies that

when

Supposing then that and using (1) and (3) we obtain


Squeeze Theorem

Suppose that holds for all in some open interval containing , except possibly at itself. Suppose also that . Then also.

Proof: From the assumptions, we know that there exists a such that and when .
These inequalities are equivalent to and when .
Using what we know about the relative ordering of , and , we have

when .

Then

when .

So

when .


Resources

Licensing

Content obtained and/or adapted from: