Calculus/Proofs of Some Basic Limit Rules

Template:Calculus/Top Nav 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.

Proof of the Constant Rule for Limits

We need to find a $\delta >0$ such that for every $\epsilon >0$ , $|b-b|<\epsilon$ whenever $0<|x-a|<\delta$ . $|b-b|=0$ and $\epsilon >0$ , so $|b-b|<\epsilon$ is satisfied independent of any value of $\delta$ ; that is, we can choose any $\delta$ we like and the $\epsilon$ condition holds.

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Proof

To prove that $\underset{x\to a}{\lim }x=a$ , we need to find a $\delta >0$ such that for every $\epsilon >0$ , $|x-a|<\epsilon$ whenever $0<|x-a|<\delta$ . Choosing $\delta =\epsilon$ satisfies this condition.

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Proof

Given the limit above, there exists in particular a $\delta >0$ such that $|f(x)-L|<\frac{\epsilon }{k}$ whenever $0<|x-a|<\delta$ , for some $k>0$ such that $|c|<k$ . Hence

$|c\cdot f(x)-c\cdot L|=|c|\cdot |f(x)-L|<k\cdot \frac{\epsilon }{k}=\epsilon$

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Proof

Since we are given that $\underset{x\to c}{\lim }f(x)=L$ and $\underset{x\to c}{\lim }g(x)=M$ , there must be functions, call them ${\delta }_f(\epsilon )$ and ${\delta }_g(\epsilon )$ , such that for all $\epsilon >0$ , $|f(x)-L|<\epsilon$ whenever $|x-c|<{\delta }_f(\epsilon )$ , and $|g(x)-M|<\epsilon$ whenever $|x-c|<{\delta }_g(\epsilon )$ .
Adding the two inequalities gives $|f(x)-L|+|g(x)-M|<2\epsilon$ . By the triangle inequality we have $|(f(x)-L)+(g(x)-M)|=|(f(x)+g(x))-(L+M)|\le |f(x)-L|+|g(x)-M|$ , so we have $|(f(x)+g(x))-(L+M)|<2\epsilon$ whenever $|x-c|<{\delta }_f(\epsilon )$ and $|x-c|<{\delta }_g(\epsilon )$ . Let ${\delta }_{fg}(\epsilon )$ be the smaller of ${\delta }_f(\frac{\epsilon }{2})$ and ${\delta }_g(\frac{\epsilon }{2})$ . Then this $\delta$ satisfies the definition of a limit for $\underset{x\to c}{\lim }[f(x)+g(x)]$ having limit $L+M$ .

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Proof

Define $h(x)=-g(x)$ . By the Scalar Product Rule for Limits, $\underset{x\to c}{\lim }h(x)=-M$ . Then by the Sum Rule for Limits, $\underset{x\to c}{\lim }[f(x)-g(x)]=\underset{x\to c}{\lim }[f(x)+h(x)]=L-M$ .

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Proof

Let $\epsilon$ be any positive number. The assumptions imply the existence of the positive numbers ${\delta }_1,{\delta }_2,{\delta }_3$ such that

$(1)|f(x)-L|<\frac{\epsilon }{2(1+|M|)}$ when $0<|x-c|<{\delta }_1$

$(2)|g(x)-M|<\frac{\epsilon }{2(1+|L|)}$ when $0<|x-c|<{\delta }_2$

$(3)|g(x)-M|<1$ when $0<|x-c|<{\delta }_3$

According to the condition (3) we see that

$|g(x)|=|g(x)-M+M|\le |g(x)-M|+|M|<1+|M|$ 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

$\begin{array}{cc}|f(x)g(x)-LM|& =|f(x)g(x)-Lg(x)+Lg(x)-LM|\\ & \le |f(x)g(x)-Lg(x)|+|Lg(x)-LM|\\ & =|g(x)|\cdot |f(x)-L|+|L|\cdot |g(x)-M|\\ & <(1+|M|)\frac{\epsilon }{2(1+|M|)}+(1+|L|)\frac{\epsilon }{2(1+|L|)}\\ & =\epsilon \end{array}$

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Proof

If we can show that $\underset{x\to c}{\lim }\frac{1}{g(x)}=\frac{1}{M}$ , 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 $\underset{x\to c}{\lim }\frac{1}{g(x)}=\frac{1}{M}$ .

Let $\epsilon$ be any positive number. The assumptions imply the existence of the positive numbers ${\delta }_1,{\delta }_2$ such that

$(1)|g(x)-M|<\frac{\epsilon |M{|}^2}{2}$ when $0<|x-c|<{\delta }_1$

$(2)|g(x)-M|<\frac{|M|}{2}$ when $0<|x-c|<{\delta }_2$

According to the condition (2) we see that

$|M|=|M-g(x)+g(x)|\le |M-g(x)|+|g(x)|<\frac{|M|}{2}+|g(x)|$ so $|g(x)|>\frac{|M|}{2}$ when $0<|x-c|<{\delta }_2$

which implies that

$(3)\left|\frac{1}{g(x)}\right|<\frac{2}{|M|}$ when $0<|x-c|<{\delta }_2$

Supposing then that $0<|x-c|<\min \{{\delta }_1,{\delta }_2\}$ and using (1) and (3) we obtain

$\begin{array}{cc}|\frac{1}{g(x)}-\frac{1}{M}|& =\left|\frac{M-g(x)}{Mg(x)}\right|\\ & =\left|\frac{g(x)-M}{Mg(x)}\right|\\ & =\left|\frac{1}{g(x)}\right|\cdot \left|\frac{1}{M}\right|\cdot |g(x)-M|\\ & <\frac{2}{|M|}\cdot \frac{1}{|M|}\cdot |g(x)-M|\\ & <\frac{2}{|M|}\cdot \frac{1}{|M|}\cdot \frac{\epsilon |M{|}^2}{2}\\ & =\epsilon \end{array}$

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Proof

From the assumptions, we know that there exists a $\delta$ such that $|g(x)-L|<\epsilon$ and $|h(x)-L|<\epsilon$ when $0<|x-c|<\delta$ .
These inequalities are equivalent to $L-\epsilon <g(x)<L+\epsilon$ and $L-\epsilon <h(x)<L+\epsilon$ when $0<|x-c|<\delta$.
Using what we know about the relative ordering of $f(x),g(x)$ , and $h(x)$ , we have
$L-\epsilon <g(x)\le f(x)\le h(x)<L+\epsilon$ when $0<|x-c|<\delta$ .
Then
$-\epsilon <f(x)-L<\epsilon$ when $0<|x-c|<\delta$ .
So
$|f(x)-L|<\epsilon$ when $0<|x-c|<\delta$ .