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 Constant Function Rule says that if $f(x)$ is identically equal to $c$, then $f(x)$ is close to $c$ for all $x$ close to $a$. This is true because $f(x)=c$ is always close to $c$.
The Identity Function Rule merely says that $f(x)=x$ is close to $a$ whenever $x$ is close to $a$.
The illustrations of these two rules are shown in Figure 1.
(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 $\lim_{x\to a}x=a$.
Figure 1
Although the above limits are obvious from the intuitive viewpoint, we can prove them rigorously using the $\epsilon$–$\delta$ definition.
Show the Proof of Theorem 1
Hide the Proof of Theorem 1
(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
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:
The limit of a constant times a function is the constant times the limit of the function.
The limit of a sum is the sum of the limits.
The limit of a difference is the difference of the limits.
The limit of a product is the product of the limits.
The limit of a quotient is the quotient of the limits provided that the limit of the denominator is not zero.
Note that the Constant Multiple Rule (1) is a special case of the Product Rule (4) when $g$ is a constant function, namely $g(x)=k$.
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,
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.
Using the $\epsilon$–$\delta$ definition, we can prove this theorem with mathematical rigor.
The above theorem can be extended to the sum, difference, or product of any finite number of functions. Namely if
\[\lim_{x\to a}f_{1}(x)=L_{1},\ \lim_{x\to a}f_{2}(x)=L_{2},\ \dots,\lim_{x\to f_{n}(x)}=L_{n}\]
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
\[\lim_{x\to a}\left[f(x)\right]^{n}=L^{n}.\]
Example 1
The graphs of $f$ and $g$ are shown in the following figure. Use the limit laws and evaluate the following limits, if they exist.
(c) From the graphs of $f$ and $g$, we see $\lim_{x\to0}f(x)=-2$, but $\lim_{x\to0}g(x)$ does not exist, because the left and right limits are not the same
As $\lim_{x\to1}g(x)$ does not exist, following the previous example, we may be tempted to say that none of the above limits exist. However, we will see that only the first one does not exist.
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
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
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
\[\lim_{x\to a}f(x)=\lim_{x\to a}g(x).\]
To be more precise, we can say if $f(x)=g(x)$ in a deleted neighborhood of $a$ 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 $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=L$.
As discussed above, when evaluating the limit, we are concerned about the values of the variable (here $t$) near $-1$ but not equal to $-1$. Thus we have,
Here we cannot simply use Limits of Rational Functions (Theorem 3) and substitute 2 for $x$ in the given expression because $\lim_{x\to2}(3x-6)=0$. However, factoring the numerator and the denominator, we obtain
The substitution of $x=-3$ in the fraction produces the meaningless fraction $0/0$. So we cannot apply Limits of Rational Functions (Theorem 3). However, factoring the numerator, we obtain
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 is also called the Squeeze Theorem or the Pinching Theorem.
The Sandwich Theorem also holds for the one-sided limits; that is, we can replace $x\to a$ by $x\to a^{-}$ or $x\to a^{+}$ in the above equations, and the theorem will be still valid.
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 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 $\lim_{x\to a}f(x)=L$.
The rigorous proof is as follows.
Read the Proof of the Sandwich Theorem
Hide the Proof
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 $x$
[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
The graph of $y=\sin x\cdot\sin(1/x)$ is shown below.
Limits Involving sin(x)/x
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:
\[
\lim_{x\to0}\frac{\sin x}{x}=1.
\]
To prove this theorem, we show that the right-hand and left-hand limits are both 1, and consequently, the limit is 1.
Show the Proof
Hide the Proof
Consider the construction shown in the following figure, where the radius of the circle is 1 ($OP=OA=1$), and $0<x<\pi/2$.
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 $\triangle OAT$, $\tan x=\frac{AT}{OA}=\frac{AT}{1}=AT$.
Let’s find each area in terms of $x$:
\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,
\[\text{area of }\triangle OAP<\text{area of sector }OAP<\text{area of }\triangle OAT\]
\[
\Rightarrow \quad \frac{1}{2}\sin x<\frac{1}{2}x<\frac{1}{2}\tan x
\]
or
\[
\sin x<x<\frac{\sin x}{\cos x}.
\]
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:
\[
1<\frac{x}{\sin x}<\frac{1}{\cos x}.
\]
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
\[
\lim_{x\to0}\frac{\sin x}{x}=1.\qquad\blacksquare
\]
Note that
\[\lim_{x\to0}\frac{x}{\sin x}=1\]
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*}
Example 15
Evaluate the following limit
\[\lim_{x\to0}\frac{\sin(\sqrt{2}x)}{x}\]
Solution
Let’s multiply the numerator and the denominator by $\sqrt{2}$
We cannot simply substitute $x=0$ in the given fraction because we will get the meaningless fraction $0/0$. Instead we use the definition of $\tan x=\sin x/\cos x$
