Calculus/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 $a,b$ are constants then $\lim _{x\to a}b=b$ .

Proof of the Constant Rule for Limits

We need to find a $\delta >0$ such that for every $\varepsilon >0$ , $|b-b|<\varepsilon$ whenever $0<|x-a|<\delta$ . $|b-b|=0$ 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 $a$ is a constant then $\lim _{x\to a}x=a$ .

Proof

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

Scalar Product Rule for Limits

Suppose that $\lim _{x\to a}f(x)=L$ for finite $L$ and that $c$ is constant. Then $\lim _{x\to a}c\cdot f(x)=c\cdot \lim _{x\to a}f(x)=c\cdot L$

Proof

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

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

Sum Rule for Limits

Suppose that $\lim _{x\to c}f(x)=L$ 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

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 ${\bigg |}(f(x)-L)+(g(x)-M){\bigg |}={\bigg |}(f(x)+g(x))-(L+M){\bigg |}\leq {\Big |}f(x)-L{\Big |}+{\Big |}g(x)-M{\Big |}$ , 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

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

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

${\Big |}g(x){\Big |}={\bigg |}g(x)-M+M{\bigg |}\leq {\Big |}g(x)-M{\Big |}+|M|<1+|M|$ 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

${\begin{aligned}{\bigg |}f(x)g(x)-LM{\bigg |}&={\bigg |}f(x)g(x)-Lg(x)+Lg(x)-LM{\bigg |}\\&\leq {\bigg |}f(x)g(x)-Lg(x){\bigg |}+{\bigg |}Lg(x)-LM{\bigg |}\\&={\Big |}g(x){\Big |}\cdot {\Big |}f(x)-L{\Big |}+|L|\cdot {\Big |}g(x)-M{\Big |}\\&<(1+|M|){\frac {\varepsilon }{2(1+|M|)}}+(1+|L|){\frac {\varepsilon }{2(1+|L|)}}\\&=\varepsilon \end{aligned}}$

Quotient Rule for Limits

Suppose that $\lim _{x\to c}f(x)=L$ and $\lim _{x\to c}g(x)=M$ and $M\neq 0$ . 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

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

$(1)\qquad {\Big |}g(x)-M{\Big |}<{\frac {\varepsilon |M|^{2}}{2}}$ when $0<|x-c|<\delta _{1}$

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

According to the condition (2) we see that

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

which implies that

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

Supposing then that $0<|x-c|<\min\{\delta _{1},\delta _{2}\}$ and using (1) and (3) we obtain

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

Theorem: (Squeeze Theorem)

Suppose that $g(x)\leq f(x)\leq h(x)$ 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

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
$L-\varepsilon <g(x)\leq f(x)\leq h(x)<L+\varepsilon$ when $0<|x-c|<\delta$ .
Then
$-\varepsilon <f(x)-L<\varepsilon$ when $0<|x-c|<\delta$ .
So
${\Big |}f(x)-L{\Big |}<\varepsilon$ when $0<|x-c|<\delta$ .