# Calculus/Proofs of Some Basic Limit Rules

 ← Formal Definition of the Limit Calculus Limits/Exercises → 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 ${\displaystyle a,b}$ are constants then ${\displaystyle \lim _{x\to a}b=b}$ .

Proof of the Constant Rule for Limits

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

Identity Rule for Limits

If ${\displaystyle a}$ is a constant then ${\displaystyle \lim _{x\to a}x=a}$ .

Proof

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

Scalar Product Rule for Limits

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

Proof

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

${\displaystyle {\Big |}c\cdot f(x)-c\cdot L{\Big |}=|c|\cdot {\Big |}f(x)-L{\Big |}

Sum Rule for Limits

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

${\displaystyle \lim _{x\to c}{\Big [}f(x)+g(x){\Big ]}=\lim _{x\to c}f(x)+\lim _{x\to c}g(x)=L+M}$
Proof

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

Difference Rule for Limits

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

${\displaystyle \lim _{x\to c}{\Big [}f(x)-g(x){\Big ]}=\lim _{x\to c}f(x)-\lim _{x\to c}g(x)=L-M}$
Proof

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

Product Rule for Limits

Suppose that ${\displaystyle \lim _{x\to c}f(x)=L}$ and ${\displaystyle \lim _{x\to c}g(x)=M}$ . Then ${\displaystyle \lim _{x\to c}{\Big [}f(x)\cdot g(x){\Big ]}=\lim _{x\to c}f(x)\cdot \lim _{x\to c}g(x)=L\cdot M}$

Proof

Let ${\displaystyle \varepsilon }$ be any positive number. The assumptions imply the existence of the positive numbers ${\displaystyle \delta _{1},\delta _{2},\delta _{3}}$ such that

${\displaystyle (1)\qquad {\Big |}f(x)-L{\Big |}<{\frac {\varepsilon }{2(1+|M|)}}}$ when ${\displaystyle 0<|x-c|<\delta _{1}}$
${\displaystyle (2)\qquad {\Big |}g(x)-M{\Big |}<{\frac {\varepsilon }{2(1+|L|)}}}$ when ${\displaystyle 0<|x-c|<\delta _{2}}$
${\displaystyle (3)\qquad {\Big |}g(x)-M{\Big |}<1}$ when ${\displaystyle 0<|x-c|<\delta _{3}}$

According to the condition (3) we see that

${\displaystyle {\Big |}g(x){\Big |}={\bigg |}g(x)-M+M{\bigg |}\leq {\Big |}g(x)-M{\Big |}+|M|<1+|M|}$ when ${\displaystyle 0<|x-c|<\delta _{3}}$

Supposing then that ${\displaystyle 0<|x-c|<\min\{\delta _{1},\delta _{2},\delta _{3}\}}$ 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}}}

Quotient Rule for Limits

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

Proof

If we can show that ${\displaystyle \lim _{x\to c}{\frac {1}{g(x)}}={\frac {1}{M}}}$ , then we can define a function, ${\displaystyle h(x)}$ as ${\displaystyle h(x)={\frac {1}{g(x)}}}$ and appeal to the Product Rule for Limits to prove the theorem. So we just need to prove that ${\displaystyle \lim _{x\to c}{\frac {1}{g(x)}}={\frac {1}{M}}}$ .

Let ${\displaystyle \varepsilon }$ be any positive number. The assumptions imply the existence of the positive numbers ${\displaystyle \delta _{1},\delta _{2}}$ such that

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

According to the condition (2) we see that

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

which implies that

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

Supposing then that ${\displaystyle 0<|x-c|<\min\{\delta _{1},\delta _{2}\}}$ 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}}}

Theorem: (Squeeze Theorem)

Suppose that ${\displaystyle g(x)\leq f(x)\leq h(x)}$ holds for all ${\displaystyle x}$ in some open interval containing ${\displaystyle c}$ , except possibly at ${\displaystyle x=c}$ itself. Suppose also that ${\displaystyle \lim _{x\to c}g(x)=\lim _{x\to c}h(x)=L}$ . Then ${\displaystyle \lim _{x\to c}f(x)=L}$ also.

Proof

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

 ← Formal Definition of the Limit Calculus Limits/Exercises → Proofs of Some Basic Limit Rules