Estimating the limit of a function using the graphical approach may not be very accurate, and as we saw in Example 4 of Section 4.1, the numerical approach may lead to incorrect results. In this section, we discuss how we can evaluate limits analytically.
In this section, we learn:
Table of Contents
Algebraic Operations on Limits
We first begin with some useful limits:
Theorem 1: 1. (Constant function rule): If $c$ is a constant, then for any number $a$ \[\lim_{x\to a}c=c\]
2. (Identity function rule): \[{\displaystyle {\lim_{x\to a}x=a.}}\]
The illustrations of these two rules are shown in Figure 1. Figure 1 Although the above limits are obvious from the intuitive viewpoint, we can prove them rigorously using the $\epsilon$–$\delta$ definition. (1) Constant function rule: Here $f(x)=c$ for every $x$ and $L=c$. We must prove that for every $\epsilon>0$, there exists a $\delta>0$ such that \[0<|x-a|<\delta\implies|c-c|<\epsilon\]
Because $|c-c|=0$, we always have \[|c-c|<\epsilon.\]
So we can choose any positive number for $\delta$, and get \[|c-c|<\epsilon\quad\text{whenever}\quad0<|x-a|<\delta\]
(2) Identity function rule: Here $f(x)=x$ and $L=a$. We must prove that for every $\epsilon>0$, there exists a $\delta>0$ such that \[0<|x-a|<\delta\implies|x-a|<\epsilon\]
Obviously if we choose $\delta\leq\epsilon$, then \[0<|x-a|<\delta\implies|x-a|<\delta\leq\epsilon.\]
We can evaluate many limits by applying the following limit laws. \[\lim_{x\to a}f(x)=L,\qquad\text{and}\qquad\lim_{x\to a}g(x)=M\]
then 1. Constant multiple rule: \[\lim_{x\to a}[kf(x)]=kL.\]
2. Sum rule: \[\lim_{x\to a}[f(x)+g(x)]=L+M\]
3. Difference rule: \[\lim_{x\to a}[f(x)-g(x)]=L-M\]
4. Product rule: \[\lim_{x\to a}[f(x)g(x)]=LM\]
5. Quotient rule: \[\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{L}{M}\quad(\text{provided }M\neq0)\]
Similar results hold for left and right hand limits; that is, we can replace $x\to a$ by $x\to a^{-}$ or $x\to a^{+}$ in the above equations. This theorem merely says: This theorem comes from the common sense that if the number $u_{1}$ is close to the number $v_{1}$ and the number $u_{2}$ is close to the number $v_{2}$ then $u_{1}\pm u_{2}$ will be close to $v_{1}\pm v_{2}$, $u_{1}u_{2}$ will be close to $v_{1}v_{2}$, and $u_{1}/u_{2}$ will be close to $v_{1}/v_{2}$. For example, \begin{align*} So it is easy to believe if $f(x)$ is close to $L$ and $g(x)$ is close to $M$ when $x$ is close to $a$, then $f(x)+g(x)$ is close to $L+M$, and $f(x)g(x)$ is close to $LM$ when $x$ is close to $a$. Similarly $f(x)/g(x)$ will be close to $L/M$ if $M\neq0$. In the Quotient Rule, we have to exclude the case of $M=0$ because division by zero is not defined. The graphs of $f$ and $g$ are shown in the following figure. Use the limit laws and evaluate the following limits, if they exist. (a) ${\displaystyle {\lim_{x\to-3}[3f(x)-2g(x)]}}$ (b) ${\displaystyle {\lim_{x\to-3}\frac{f(x)}{4g(x)}}}$ (c) ${\displaystyle {\lim_{x\to0}[f(x)g(x)]}}$
The graphs of $f$ and $g$ are shown in the following figure. Use the limit laws and evaluate the following limits, if they exist. (a) ${\displaystyle \lim_{x\to1}[f(x)-g(x)]}$ (b) ${\displaystyle \lim_{x\to1}[f(x)g(x)]}$ (c) ${\displaystyle \lim_{x\to1}\frac{g(x)}{f(x)}}$
Show that ${\displaystyle \lim_{x\to0}|x|}=0$.
In general: Theorem 3. (Limits of Polynomials and Rational Functions): Let $P(x)$ and $Q(x)$ be two polynomials and $a$ be a real number. Then 1. Polynomial functions: \[\lim_{x\to a}P(x)=P(a)\]
2. Rational functions: \[\lim_{x\to a}\frac{P(x)}{Q(x)}=\frac{P(a)}{Q(a)}\quad(\text{provided }Q(a)\neq0)\]
Similar results hold for left and right hand limits; that is, we can replace $x\to a$ by $x\to a^{-}$ or $x\to a^{+}$ in the above equations. (a) For the left hand limit as $t\to-1^{-}$, we use the equation $g(t)=2t^{2}+1$ \[\lim_{t\to-1^{-}}g(t)=\lim_{t\to-1^{-}}(2t^{2}+1)=2(-1)^{2}+1=3.\]
(b) For the right hand limit as $t\to-1^{+1}$, we use the equation $g(t)=4+t$ \[\lim_{t\to-1^{+}}g(t)=\lim_{t\to-1^{+}}(4+t)=4+(-1)=3.\]
(c) Because ${\displaystyle \lim_{t\to-1^{-}}g(t)=\lim_{t\to-1^{+}}g(t)=3}$, by Theorem 1 in Section 4.3, \[\lim_{t\to-1}g(t)=3.\]
The graph of $y=g(t)$ is shown in the following figure. In an example in Section 4.2, we showed that \[\lim_{x\to a}\sqrt{x}=\sqrt{a},\qquad(a>0)\]
Similarly, using the $\epsilon$-$\delta$ definition, we can show that the following theorem is true. This limit is consistent with the appearance of the graph $y=\sqrt[n]{x}$. Theorem 4: If $n$ is a positive integer, then \[\lim_{x\to a}\sqrt[n]{x}=\sqrt[n]{a}\]
[If $n$ is even, we assume that $a>0$.]
In general, we have the following theorem which will be shown in the Section on Continuity that it is a consequence of the above theorem. Theorem 5 (Root Rule): If $n$ is a positive integer, then \[\lim_{x\to a}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x\to a}f(x)}\]
[If $n$ is even we assume that $\lim_{x\to a}f(x)>0$]
To show that the above theorem is plausible, assume that $\lim_{x\to a}\sqrt[n]{f(x)}=L$ exists. Then by the general form of the Product Rule, \begin{align*} \[\Rightarrow L=\sqrt[n]{\lim_{x\to a }f(x)}.\]
Suppose we want to evaluate \[\lim_{x\to1}\frac{x^{2}-1}{x-1}\]
Because the denominator $x-1$ is zero at $x=1$, we cannot use the above theorem. So what can we do here? Let\textquoteright s simplify this fraction \[\frac{x^{2}-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1\quad(\text{if }x\neq1)\]
This shows us that two functions $y=\frac{x^{2}-1}{x-1}$ and $y=x+1$ are identical except when $x=1$. But the limit of $\frac{x^{2}-1}{x-1}$ depends only on the values of this function for $x$ near $1$ and the value of a function at $x=1$ does not influence the limit. So we must have \[\lim_{x\to1}\frac{x^{2}-1}{x-1}=\lim_{x\to1}x+1\]
By Theorem 3: \[\lim_{x\to1}(x+1)=1+1=2\]
Therefore: \[\lim_{x\to1}\frac{x^{2}-1}{x-1}=2.\]
In general recall that $\lim_{x\to a}f(x)$ depends only on the values of $f(x)$ for $x$ close to $a$ and not on the value of $f$ at $x=a$ nor on the values of $f(x)$ for $x$ far away from $a$. So if there is a function $g$ such that $f(x)=g(x)$ for $x$ close to $a$ but not necessarily for $x=a$, then \[\lim_{x\to a}f(x)=\lim_{x\to a}g(x)\]
Therefore, we have the following theorem: Theorem 6 (Replacement Rule): If \[f(x)=g(x)\]
for all $x$ close to $a$, but not necessarily including $x=a$, then \[\lim_{x\to a}f(x)=\lim_{x\to a}g(x).\]
Illustration of this theorem is presented in the following figure. We can see that $f(x)=g(x)$ for all $x$ between $b$ and $c$ (or in an open interval containing $a$) except for $x=a$. Therefore $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=L$. The following theorem helps us find a variety of limits: Theorem 7 (The Sandwich Theorem): If we have \[g(x)\leq f(x)\leq h(x)\]
for all $x$ in some open interval containing $a$ except possibly at $x=a$ itself and if \[\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L,\]
then we also have \[\lim_{x\to a}f(x)=L.\]
The Sandwich Theorem states that if a function $f$ is sandwiched between two functions $g$ and $h$ near $a$ and if $g$ and $h$ approach the same number $L$, then $f$ must approach $L$ too (as $x\to a$, where else can $f$ go to, if not to $L$?). The validity of this theorem is suggested by the following figure. The rigorous proof is as follows. Suppose $\epsilon>0$ is given. We need to show that there exists a $\delta>0$ such that for all $x$ \[0<|x-a|<\delta\Rightarrow|f(x)-L|<\epsilon.\]
Because $\lim_{x\to a}g(x)=L$ and $\lim_{x\to a}h(x)=L$, there exist some $\delta_{1}>0$ and $\delta_{2}>0$ such that for all $x$ \[0<|x-a|<\delta_{1}\Rightarrow L-\epsilon<g(x)<L+\epsilon\]
and \[0<|x-a|<\delta_{2}\Rightarrow L-\epsilon<h(x)<L+\epsilon\]
[Note that $|g(x)-L|<\epsilon$ is equivalent to $-\epsilon<g(x)-L<\epsilon$ or if we add $L$ to each side it is equivalent to $L-\epsilon<g(x)<L+\epsilon$. See Property 7(i) in the Section on Absolute Value]
Now if we choose $\delta=\min\{\delta_{1},\delta_{2}\}$, then for all $x$ if $0<|x-a|<\delta$ then \[L-\epsilon<h(x)\leq f(x)\leq g(x)<L+\epsilon\]
That is, \[L-\epsilon<f(x)<L+\epsilon\]
or \[|f(x)-L|<\epsilon.\]
For every number $t$, we know \[t-1<\left\lfloor t\right\rfloor \leq t.\]
Therefore, for every $x\neq0$ \[\frac{1}{x}-1<\left\lfloor \frac{1}{x}\right\rfloor \leq\frac{1}{x}.\]
If $x>0$, we can multiply each side by $x$ and the directions of inequalities are preserved: \[1-x<x\left\lfloor \frac{1}{x}\right\rfloor \leq1.\]
Because $\lim_{x\to0}(1-x)=1-0=1$, by the Sandwich Theorem: \[\lim_{x\to0^{+}}x\left\lfloor \frac{1}{x}\right\rfloor =1.\]
If $x<0$, we still multiply both sides by $x$ but the directions of inequalities are reversed: \[1-x>x\left\lfloor \frac{1}{x}\right\rfloor \geq1.\]
Again because $\lim_{x\to0^{-}}(1-x)=1-0=1$, by the Sandwich Theorem \[\lim_{x\to0^{-}}x\left\lfloor \frac{1}{x}\right\rfloor =1.\]
Because $\lim_{x\to0^{-}}x\left\lfloor \frac{1}{x}\right\rfloor =\lim_{x\to0^{+}}x\left\lfloor \frac{1}{x}\right\rfloor =1$, we conclude that \[\lim_{x\to0}x\left\lfloor \frac{1}{x}\right\rfloor =0.\]
The graphs of $y=x\left\lfloor \frac{1}{x}\right\rfloor $, $y=1-x$, and $y=1$ are depicted in the following figure. In the Section on the Trigonometric Inequalities, we learned that for all values of $x$, we have \[-|x|\leq\sin x\leq|x|\]
and \[-|x|\leq1-\cos x\leq|x|.\]
We can use the above inequalities and the Sandwich Theorem to solve the next example.
(a) Note that for $x\neq0$, we have \[-1\leq\sin\frac{1}{x}\leq1.\tag{i}\]
If $x>0$ and we multiply each side of (i) by $x$, the directions of the inequalities will not change. That is, \[-x\leq x\sin\frac{1}{x}\leq x.\qquad(x>0)\]
Because \[\lim_{x\to0^{+}}x=\lim_{x\to0^{+}}(-x)=0\]
it follows from the Sandwich Theorem that \[\lim_{x\to0^{+}}x\sin\frac{1}{x}=0.\]
If $x<0$ and we multiply each side of (i) by $x$, we need to reverse the direction of the inequalities. That is, \[-x\geq x\sin\frac{1}{x}\geq x.\qquad(x<0)\]
Again because \[\lim_{x\to0^{-}}x=\lim_{x\to0^{-}}(-x)=0\]
it follows from the Sandwitch Theorem that \[\lim_{x\to0^{-}}x\sin\frac{1}{x}=0.\]
Because the left and right limits are equal, we conclude that \[\lim_{x\to0}x\sin\frac{1}{x}=0.\]
The graph of $y=x\sin(1/x)$ is shown below (b) Similar to part (a), we start from the fact that \[-1\leq\cos t\leq1\]
and if we replace $t$ with $1/x^{2}$ when $x\neq0$, we get \[-1\leq\cos\left(\frac{1}{x^{2}}\right)\leq1.\]
Multiplying each side by $\sqrt{x}$ \[-\sqrt{x}\leq\sqrt{x}\cos\left(\frac{1}{x^{2}}\right)\leq\sqrt{x}.\]
Because \[\lim_{x\to0^{+}}\sqrt{x}=\lim_{x\to0^{+}}(-\sqrt{x})=0\]
by the Sandwitch theorem \[\lim_{x\to0^{+}}\sqrt{x}\cos\left(\frac{1}{x^{2}}\right)=0.\]
The graph of $y=\sqrt{x}\cos(1/x^{2})$ is shown below. (c) Similar to part (a), we start with \[-1\leq\sin\left(\frac{1}{x}\right)\leq1\qquad(\text{if }x\neq0).\]
If $0<x<\pi$, then $\sin x>0$ and therefore \[-\sin x\leq\sin x\sin\left(\frac{1}{x}\right)\leq\sin x\qquad(0<x<\pi)\]
and if $-\pi<x<0$, then $\sin x<0$ and therefore \[-\sin x\geq\sin x\sin\left(\frac{1}{x}\right)\geq\sin x.\qquad(-\pi<x<0)\]
Because \[\lim_{x\to0}\sin x=\lim_{x\to0}(-\sin x)=0\]
it follows from the Sandwich Theorem that \[\lim_{x\to0}\sin x\sin\frac{1}{x}=0.\]
The graph of $y=\sin x\cdot\sin(1/x)$ is shown below. Recall that when there is no $^{\circ}$ notation, the default is that $x$ is measured in radian. In Example 3 in the Section on the Concept of a Limit [you need to click on “Show Some Examples” to be able to see this example], we saw that $\lim_{x\to0}\sin x/x=1$. This limit is of importance and we can solve many similar exercises using this limit. In this section, we prove that $\lim_{x\to0}\sin x/x=1$ using the Sandwich Theorem. Theorem 8: Consider the construction shown in the following figure, where the radius of the circle is 1 ($OP=OA=1$), and $0<x<\pi/2$.
Let’s find each area in terms of $x$: \[\text{area of }\triangle OAP<\text{area of sector }OAP<\text{area of }\triangle OAT\]
\[ Because $0<x<\pi/2$, the sine of $x$ is positive and dividing each term by $\sin x$ does not change the directions of inequalities:
In general for a constant $A$ \[ \bbox[#F2F2F2,5px,border:2px solid black]{\lim_{x\to0}\frac{\sin Ax}{x}=A.}\] \[ \bbox[#F2F2F2,5px,border:2px solid black]{\lim_{x\to 0}\frac{\tan x}{x}=1}\] Another importance limit is \[\lim_{x\to0}\frac{1-\cos x}{x^{2}}\]
Notice that we cannot simply substitute $x=0$ in the given expression because we get the meaningless fraction $0/0$ upon substitution. Recall that the half-angle formula: \[\frac{1-\cos2\theta}{2}=\sin^{2}\theta\]
Now let $x=2\theta$. Therefore, \[1-\cos x=2\sin^{2}\left(\frac{x}{2}\right)\]
and \begin{align*} We just saw that if $A$ is a constant \[\lim_{x\to0}\frac{\sin(Ax)}{x}=A\]
Therefore \[\lim_{x\to0}\frac{1-\cos x}{x^{2}}=2\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{2}.\]
\[ \bbox[#F2F2F2,5px,border:2px solid black]{\lim_{x\to0}\frac{1-\cos x}{x^{2}}=\frac{1}{2}.}\] From the above limit, we can conclude that \[ \bbox[#F2F2F2,5px,border:2px solid black]{\lim_{x\to0}\frac{1-\cos x}{x}=0}\] because \begin{align*}
(a) Graph of a constant function. It is clear that $\lim_{x\to a}c=c$
(b) Graph of the identity function $y=x$. It is clear that $\lim_{x\to a}x=a$.
Show the Proof of Theorem 1
Hide the Proof of Theorem 1
Basic Algebraic Operations
7.01+11.001 & \approx7+11\\
7.01-11.001 & \approx7-11\\
7.01\times11.001 & \approx7\times11\\
7.01/11.001 & \approx7/11
\end{align*}
\[\lim_{x\to a}f_{1}(x)=L_{1},\ \lim_{x\to a}f_{2}(x)=L_{2},\ \dots,\lim_{x\to f_{n}(x)}=L_{n}\]
then
\[\lim_{x\to a}[f_{1}(x)\pm f_{2}(x)\pm\cdots\pm f_{n}(x)]=L_{1}\pm L_{2}\pm\cdots\pm L_{n}\]
and
\[\lim_{x\to a}[f_{1}(x)f_{2}(x)\cdots f_{n}(x)]=L_{1}L_{2}\cdots L_{n}\]
Specifically, if $n$ is a positive integer then
\[\lim_{x\to a}\left[f(x)\right]^{n}=L^{n}.\]
Limits of Polynomial and Rational Functions
Root Rule
\underbrace{{\left(\lim_{x\to a}\sqrt[n]{f(x)}\right)\cdots\left(\lim_{x\to a}\sqrt[n]{f(x)}\right)}}_{n\text{ times}}&=\lim_{x\to a}\left(\underbrace{\sqrt[n]{f(x)}\cdots \sqrt[n]{f(x)}}_{n\text{ times}}\right)\\
\underbrace{L\cdots L}_{n\text{ times}}&=\lim_{x\to a}{\Bigg(}f(x){\Bigg)}
\end{align*} Replacement Rule
The Sandwich Theorem
The Sandwich Theorem. If $g(x)\leq f(x)\leq h(x)$ for all $x$ near $a$ (except possibly at $a$) and $\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L$, then $\lim_{x\to a}f(x)=L$.
Read the Proof of the Sandwich Theorem
Hide the Proof
Graphs of $y=x\lfloor\frac{1}{x}\rfloor$, $y=1-x$, and $y=1$.
Trigonometric Functions
Limits Involving sin(x)/x
\[
\lim_{x\to0}\frac{\sin x}{x}=1.
\]
Show the Proof
Hide the Proof
Notice that
\[\text{area of }\triangle OAP<\text{area of sector }OAP<\text{area of }\triangle OAT\]
In $\triangle OPH$, $\sin x=\frac{PH}{OP}=\frac{PH}{1}=PH$. In $\triangle OAT$, $\tan x=\frac{AT}{OA}=\frac{AT}{1}=AT$.
\begin{align*}
\text{area of }\triangle OAP & =\frac{1}{2}\text{base}\times\text{height}\\
& =\frac{1}{2}OA\times PH\\
& =\frac{1}{2}(1)(\sin x)\\
& =\frac{1}{2}\sin x.
\end{align*}
Because the central angle of the sector is $x$:
\begin{align*}
\text{area of sector }OAP & =\frac{1}{2}r^{2}\theta\\
& =\frac{1}{2}(1)x\\
& =\frac{x}{2}.
\end{align*}
And
\begin{align*}
\text{area of }\triangle OAT & =\frac{1}{2}\text{base}\times\text{height }\\
& =\frac{1}{2}OA\times AT\\
& =\frac{1}{2}(1)(\tan x)\\
& =\frac{1}{2}\tan x.
\end{align*}
Thus,
\Rightarrow \quad \frac{1}{2}\sin x<\frac{1}{2}x<\frac{1}{2}\tan x
\]
or
\[
\sin x<x<\frac{\sin x}{\cos x}.
\]
\[
1<\frac{x}{\sin x}<\frac{1}{\cos x}.
\]
Taking reciprocals reverses the inequalities (see property 7 of inequalities in the Section on Inequalities):
\[1>\frac{\sin x}{x}>\cos x.\]
Now we show that the last double inequality holds also for $-\pi/2<x<0$. If $-\pi/2<x<0$, then $0<-x<\pi/2$ and therefore
\[1>\frac{\sin(-x)}{(-x)}>\cos(-x).\]
Recall that $\sin x$ is an odd function and $\cos x$ is an even function. That is,
\[\sin(-x)=-\sin x,\quad\cos(-x)=\cos x\]
Thus
\[1>\frac{-\sin x}{-x}>\cos(-x)=\cos x\]
So we proved
\[1<\frac{\sin x}{x}<\cos x.\qquad(-\frac{\pi}{2}<x<\frac{\pi}{2},x\neq0)\]
Because $\lim_{x\to0}\cos x=\cos0=1$ and $\lim_{x\to0}1=1$, the Sandwich Theorem gives
\[
\lim_{x\to0}\frac{\sin x}{x}=1.\qquad\blacksquare
\]
\[\lim_{x\to0}\frac{x}{\sin x}=1\]
because
\begin{align*}
\lim_{x\to0}\frac{x}{\sin x} & =\lim_{x\to0}\frac{1}{\dfrac{\sin x}{x}}\\
& =\frac{\lim_{x\to0}1}{\lim_{x\to0}\dfrac{\sin x}{x}}\\
& =\frac{1}{1}=1
\end{align*}
\lim_{x\to0}\frac{1-\cos x}{x^{2}} & =\lim_{x\to0}\frac{2\sin^{2}\left(\dfrac{x}{2}\right)}{x^{2}}\\
& =2\lim_{x\to0}\frac{\sin(x/2)}{x}\cdot\frac{\sin(x/2)}{x}\\
& =2\lim_{x\to0}\frac{\sin(x/2)}{x}\cdot\lim_{x\to0}\frac{\sin(x/2)}{x}
\end{align*}
\lim_{x\to0}\frac{1-\cos x}{x} & =\lim_{x\to0}\left(x\frac{1-\cos x}{x^{2}}\right)\\
& =\left(\lim_{x\to0}x\right)\left(\lim_{x\to0}\frac{1-\cos x}{x^{2}}\right)\\
& =(0)\left(\frac{1}{2}\right)=0.
\end{align*}