|
|
Line 20: |
Line 20: |
| ===Sum Rule for Limits=== | | ===Sum Rule for Limits=== |
| : Suppose that <math>\lim_{x\to c}f(x)=L</math> and <math>\lim_{x\to c}g(x)=M</math>. Then, | | : Suppose that <math>\lim_{x\to c}f(x)=L</math> and <math>\lim_{x\to c}g(x)=M</math>. Then, |
− | :: <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>}}. | + | :: <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>. |
| | | |
| 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/> | | 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/> |
Line 27: |
Line 27: |
| ===Difference Rule for Limits=== | | ===Difference Rule for Limits=== |
| : Suppose that <math>\lim_{x\to c}f(x)=L</math> and <math>\lim_{x\to c}g(x)=M</math>. Then | | : Suppose that <math>\lim_{x\to c}f(x)=L</math> and <math>\lim_{x\to c}g(x)=M</math>. Then |
− | :: <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>}} | + | :: <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> |
| | | |
| 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>. | | 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>. |
Line 33: |
Line 33: |
| ===Product Rule for Limits=== | | ===Product Rule for Limits=== |
| : Suppose that <math>\lim_{x\to c}f(x)=L</math> and <math>\lim_{x\to c}g(x)=M</math> . Then | | : Suppose that <math>\lim_{x\to c}f(x)=L</math> and <math>\lim_{x\to c}g(x)=M</math> . Then |
− | :: <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>}} | + | :: <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> |
| 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/> | | 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/> |
| :<math>(1)\qquad\Big|f(x)-L\Big|<\frac{\varepsilon}{2(1+|M|)}</math> when <math>0<|x-c|<\delta_1</math> | | :<math>(1)\qquad\Big|f(x)-L\Big|<\frac{\varepsilon}{2(1+|M|)}</math> when <math>0<|x-c|<\delta_1</math> |
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
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2245bcbbf23e62d18e911d97fb3374f21029d253)
Proof: Define
. By the Scalar Product Rule for Limits,
. Then by the Sum Rule for Limits,
.
Product Rule for Limits
- Suppose that
and
. Then
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8bb3a49fa4c6cf42d9198a0a3a589f0d37f801ef)
Proof: Let
be any positive number. The assumptions imply the existence of the positive numbers
such that
when ![{\displaystyle 0<|x-c|<\delta _{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b29ca9d8c6122845e587d966d292950b59a01d98)
when ![{\displaystyle 0<|x-c|<\delta _{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ff0b426e60094952c83322ab7b2c4e7311d2cac)
when ![{\displaystyle 0<|x-c|<\delta _{3}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bcc23e6bf8cda782c88b2a99cc4ad11eb9e338a7)
According to the condition (3) we see that
when ![{\displaystyle 0<|x-c|<\delta _{3}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bcc23e6bf8cda782c88b2a99cc4ad11eb9e338a7)
Supposing then that
and using (1) and (2) we obtain
![{\displaystyle {\begin{aligned}{\bigg |}f(x)g(x)-LM{\bigg |}&={\bigg |}f(x)g(x)-Lg(x)+Lg(x)-LM{\bigg |}\\&\leq {\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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29f61e67d3dfdf10eab3770bafb65505e7eb19a6)
Quotient Rule for Limits
- Suppose that
and
and
. Then
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0171a9a0dfe00d5c16e727eaa665c634cd3e1fbc)
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 ![{\displaystyle 0<|x-c|<\delta _{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b29ca9d8c6122845e587d966d292950b59a01d98)
when ![{\displaystyle 0<|x-c|<\delta _{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ff0b426e60094952c83322ab7b2c4e7311d2cac)
According to the condition (2) we see that
so
when ![{\displaystyle 0<|x-c|<\delta _{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ff0b426e60094952c83322ab7b2c4e7311d2cac)
which implies that
when ![{\displaystyle 0<|x-c|<\delta _{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ff0b426e60094952c83322ab7b2c4e7311d2cac)
Supposing then that
and using (1) and (3) we obtain
![{\displaystyle {\begin{aligned}\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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/320b15f6685efafb1f8b8add52f02a83c534bcab)
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: