Calculus/Proofs of Some Basic Limit Rules

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Proofs of Some Basic Limit Rules

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 b and c are constants then \lim_{x\to c}b=b.

Proof of the Constant Rule for Limits:
To prove that \lim_{x\to c}b=b , we need to find a \delta>0 such that for every \varepsilon>0 , |b-b|<\varepsilon whenever |x-c|<\delta . |b-b|=0 and \varepsilon>0 , so |b-b|<\varepsilon is satisfied independent of any value of \delta ; that is, we can choose any \delta we like and the \varepsilon condition holds.

Identity Rule for Limits

If c is a constant then \lim_{x\to c}x=c.

Proof of the Identity Rule for Limits:
To prove that \lim_{x\to c}x=c , we need to find a \delta>0 such that for every \varepsilon>0 , |x-c|<\varepsilon whenever |x-c|<\delta . Choosing \delta=\varepsilon satisfies this condition.

Scalar Product Rule for Limits

Suppose that \lim_{x\to c}f(x)=L for finite L and that k is constant. Then \lim_{x\to c}k\cdot f(x)=k\cdot\lim_{x\to c}f(x)=k\cdot L

Proof of the Scalar Product Rule for Limits:
Since we are given that \lim_{x\to c}f(x)=L, there must be some function, call it \delta_{f}(\varepsilon), such that for every \varepsilon>0 , |f(x)-L|<\varepsilon whenever |x-c|<\delta_{f}(\varepsilon) . Now we need to find a \delta_{kf}(\varepsilon) such that for all \varepsilon>0 , \Big|k\cdot f(x)-k\cdot L\Big|<\varepsilon whenever |x-c|<\delta_{kf}(\varepsilon) .
First let's suppose that k>0 . \Big|k\cdot f(x)-k\cdot L\Big|=k\Big|f(x)-L\Big|<\varepsilon , so \Big|f(x)-L\Big|<\tfrac{\varepsilon}{k} . In this case, letting \delta_{kf}(\varepsilon)=\delta_{f}(\tfrac{\varepsilon}{k}) satisfies the limit condition.
Now suppose that k=0 . Since f(x) has a limit at x=c , we know from the definition of a limit that f(x) is defined in an open interval D that contains c (except maybe at c itself) . In particular, we know that f(x) doesn't blow up to infinity within D (except maybe at c, but that won't affect the limit), so that 0\cdot f(x)=0 in D. Since k f(x) is the constant function 0 in D, the limit \lim_{x\to c}k\cdot f(x)=0 by the Constant Rule for Limits.
Finally, suppose that k<0 . \Big|k\cdot f(x)-k\cdot L\Big|=-k\Big|f(x)-L\Big|<\varepsilon , so \Big|f(x)-L\Big|<-\tfrac{\varepsilon}{k} . In this case, letting \delta_{kf}(\varepsilon)=\delta_{f}(-\tfrac{\varepsilon}{k}) satisfies the limit condition.

Sum Rule for Limits
Suppose that \lim_{x\to c}f(x)=L and \lim_{x\to c}g(x)=M . Then

\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 of the Sum Rule for Limits:
Since we are given that \lim_{x\to c}f(x)=L and \lim_{x\to c}g(x)=M , there must be functions, call them \delta_{f}(\varepsilon) and \delta_{g}(\varepsilon) , such that for all \varepsilon>0 , \Big|f(x)-L\Big|<\varepsilon whenever |x-c|<\delta_{f}(\varepsilon) , and \Big|g(x)-M\Big|<\varepsilon whenever |x-c|<\delta_{g}(\varepsilon) .
Adding the two inequalities gives \Big|f(x)-L\Big|+\Big|g(x)-M\Big|<2\varepsilon . By the triangle inequality we have \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 \bigg|(f(x)+g(x))-(L+M)\bigg|<2\varepsilon whenever |x-c|<\delta_{f}(\varepsilon) and |x-c|<\delta_{g}(\varepsilon) . Let \delta_{fg}(\varepsilon) be the smaller of \delta_{f}(\tfrac{\varepsilon}{2}) and \delta_{g}(\tfrac{\varepsilon}{2}) . Then this \delta satisfies the definition of a limit for \lim_{x\to c}\Big[f(x)+g(x)\Big] having limit L+M .

Difference Rule for Limits
Suppose that \lim_{x\to c}f(x)=L and \lim_{x\to c}g(x)=M . Then

\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 of the Difference Rule for Limits: Define h(x)=-g(x) . By the Scalar Product Rule for Limits, \lim_{x\to c}h(x)=-M . Then by the Sum Rule for Limits, \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 \lim_{x\to c}f(x)=L and \lim_{x\to c}g(x)=M . Then

\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 of the Product Rule for Limits:[1]
Let \varepsilon be any positive number. The assumptions imply the existence of the positive numbers \delta_1,\delta_2,\delta_3 such that

(1)\qquad\Big|f(x)-L\Big|<\frac{\varepsilon}{2(1+|M|)} when 0<|x-c|<\delta_1
(2)\qquad\Big|g(x)-M\Big|<\frac{\varepsilon}{2(1+|L|)} when 0<|x-c|<\delta_2
(3)\qquad\Big|g(x)-M\Big|<1 when 0<|x-c|<\delta_3

According to the condition (3) we see that

\Big|g(x)\Big|=\bigg|g(x)-M+M\bigg|\le\Big|g(x)-M\Big|+|M|<1+|M| when 0<|x-c|<\delta_3

Supposing then that 0<|x-c|<\min\{\delta_1,\delta_2,\delta_3\} and using (1) and (2) we obtain

\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 \lim_{x\to c}f(x)=L and \lim_{x\to c}g(x)=M and M\ne 0 . Then

\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 of the Quotient Rule for Limits:
If we can show that \lim_{x\to c}\frac{1}{g(x)}=\frac{1}{M}, then we can define a function, h(x) as 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 \lim_{x\to c}\frac{1}{g(x)}=\frac{1}{M} .
Let \varepsilon be any positive number. The assumptions imply the existence of the positive numbers \delta_1,\delta_2 such that

(1)\qquad\Big|g(x)-M\Big|<\varepsilon|M|(1+|M|) when 0<|x-c|<\delta_1
(2)\qquad\Big|g(x)-M\Big|<1 when 0<|x-c|<\delta_{2}

According to the condition (2) we see that

\Big|g(x)\Big|=\Big|g(x)-M+M\Big|\le\Big|g(x)-M\Big|+|M|<1+|M| when 0<|x-c|<\delta_2

which implies that

(3)\qquad\left|\frac{1}{g(x)}\right|>\frac{1}{1+|M|} when 0<|x-c|<\delta_2

Supposing then that 0<|x-c|<\min\{\delta_1,\delta_2\} and using (1) and (3) we obtain

\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{g(x)-M}{M}\right|\\
&<\frac{1}{1+|M|}\cdot\left|\frac{g(x)-M}{M}\right|\\
&<\frac{1}{1+|M|}\cdot\left|\frac{\varepsilon|M|(1+|M|)}{M}\right|\\
&=\varepsilon
\end{align}
Theorem: (Squeeze Theorem)
Suppose that g(x)\le f(x)\le h(x) holds for all x in some open interval containing c , except possibly at x=c itself. Suppose also that \lim_{x\to c}g(x)=\lim_{x\to c}h(x)=L . Then \lim_{x\to c}f(x)=L also.

Proof of the Squeeze Theorem:
From the assumptions, we know that there exists a \delta such that \Big|g(x)-L\Big|<\varepsilon and \Big|h(x)-L\Big|<\varepsilon when 0<|x-c|<\delta .
These inequalities are equivalent to L-\varepsilon<g(x)<L+\varepsilon and L-\varepsilon<h(x)<L+\varepsilon when 0<|x-c|<\delta.
Using what we know about the relative ordering of f(x),g(x) , and h(x) , we have
L-\varepsilon<g(x)<f(x)<h(x)<L+\varepsilon when 0<|x-c|<\delta .
or
-\varepsilon<g(x)-L<f(x)-L<h(x)-L<\varepsilon when 0<|x-c|<\delta .
So
\Big|f(x)-L\Big|<\max\Big\{\Big|g(x)-L\Big|,\Big|h(x)-L\Big|\Big\}<\varepsilon when 0<|x-c|<\delta .

Notes[edit]

  1. This proof is adapted from one found at planetmath.org/encyclopedia/ProofOfLimitRuleOfProduct.html due to Planet Math user pahio and made available under the terms of the Creative Commons By/Share-Alike License.
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Proofs of Some Basic Limit Rules