Calculus/Proofs of Some Basic Limit Rules

< Calculus

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

b,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

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 $\varepsilon>0$ , ${\Big |}f(x)-L{\Big |}<\varepsilon$ 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 $k\cdot f(x)$ 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 ${\bigg |}(f(x)-L)+(g(x)-M){\bigg |}={\bigg |}(f(x)+g(x))-(L+M){\bigg |}\leq {\Big |}f(x)-L{\Big |}+{\Big |}g(x)-M{\Big |}$ , 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 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 |}\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 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|\cdot {\Big |}|M|-1{\Big |}

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

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

|g(x)|>|M|-1

when

0<|x-c|<\delta _{2}

which implies that

(3)\qquad \left|{\frac {1}{g(x)}}\right|<{\frac {1}{|M|-1}}\leq {\Big |}{\frac {1}{|M|-1}}{\Big |}

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 {g(x)-M}{M}}\right|\\&<\left|{\frac {1}{|M|-1}}\right|\cdot \left|{\frac {g(x)-M}{M}}\right|\\&<\left|{\frac {1}{|M|-1}}\right|\cdot \left|{\frac {\varepsilon |M|\cdot {\Big |}|M|-1{\Big |}}{M}}\right|\\&=\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 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

↑ 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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