Limits Examples

Example 2.1 Compute the value of the limit given by,

\lim _{x \rightarrow-1}\left(x^{2}+3 x-3\right) \text {. }

Solution First, we use Properties I and II to break (2.7) into three separate limits,

\lim _{x \rightarrow-1}\left(x^{2}+3 x-3\right)=\lim _{x \rightarrow-1} x^{2}+3 \lim _{x \rightarrow-1} x-\lim _{x \rightarrow-1} 3 .

Evaluating each limit gives,

\begin{aligned} \lim _{x \rightarrow-1}\left(x^{2}+3 x-3\right) & =(-1)^{2}+3(-1)-3 \\ & =-5 \end{aligned}

An important thing to note about Example 2.1 is that we obtained the limit of $x^{2}$ and $3 x$ simply by evaluating these terms at the point $x=-1$ . This can be done in cases where the functions are, as often expressed in an ad hoc way in mathematics, 'nice'. We define 'nice' functions formally in Section 2.4 later in this chapter. At this point we simply note that for 'nice' functions, what is happening sufficiently close to the point $a$ is exactly what happens at $a$ . If $f(x)$ is a 'nice' function then,

\lim _{x \rightarrow a} f(x)=f(a)

this is true as $a$ is approached from both sides. While this result can be used for a large class of functions (all polynomials for instance), it cannot be used to evaluate all limits, especially in cases where we get an indeterminate form for the limit. The ratios $0 / 0$ or $\infty / \infty$ are common examples of indeterminate forms. As an example consider the limit,

\lim _{x \rightarrow 0} \frac{\sin (2 x)}{x}

note that the quotient rule given by Property IV cannot be used here since the limit of the denominator, i.e. $\lim _{x \rightarrow 0} x=0$ . The limit does exist however and it is equal to 2. To evaluate this limit, one can use a Maclaurin series expansion (see Chapter 6) or L'Hôpital's rule given in Section 2.3. In some cases that result in indeterminate forms, a limit that exists can be evaluated through simplification; we saw an example of this in Eq. (2.4) above. Next, we work through some examples of more difficult limits. Example 2.2 Evaluate the limit,

\lim _{x \rightarrow 1} \frac{\sqrt{2-x}-1}{1-x}

Solution This limit is of the form $0 / 0$ . One way to go about evaluating the limit is by introducing a new variable for the term in the denominator, e.g. $s=1-x$ rendering the limit in Eq. (2.8) as,

\lim _{s \rightarrow 0} \frac{\sqrt{1+s}-1}{s}

Next, we notice that we can rewrite (2.9) as

\begin{aligned} \lim _{s \rightarrow 0} \frac{(\sqrt{1+s}-1)}{s} \frac{(\sqrt{1+s}+1)}{\sqrt{1+s}+1} & =\frac{1+s-1}{s(\sqrt{1+s}+1)} \\ & =\frac{1}{\sqrt{1+s}+1} \end{aligned}

provided $s \neq 0$ . As $s \rightarrow 0$ , the limit is $1 / 2$ .

In Example 2.2 starting from Eq. (2.9), we can also make use of the binomial theorem to approximate $\sqrt{1+s}$ for small $s$ ensuring convergence. We discuss series and their convergence in Chapter 6 in more detail but here we give the definition of the general binomial theorem followed by the binomial series expansion in Definition 2.1 in order to finish off the example.

Definition 2.1 If $n$ is any positive integer then,

(a+b)^{n}=\sum_{m=0}^{n} \frac{n !}{m !(n-m) !} a^{n-m} b^{m} .

Now, if $n$ is any number (including fractions, negative integers, etc.) and $\underline{|x|<1}$ then

(1+x)^{n}=\sum_{k=0}^{\infty}\left(\begin{array}{l} n \\ k \end{array}\right) x^{k}

where

\left(\begin{array}{l} n \\ k \end{array}\right)=\frac{n(n-1)(n-2) \cdots(n-k+1)}{k !} .

For the case where $n$ is not a positive integer, the binomial series given by Eq. (2.12) does not terminate and the series converges for $-1<x<1$ . Note that (2.12) cannot be applied to a series of the form $(a+x)^{n}$ directly; instead, we need to rewrite it as

(a+x)^{n}=a^{n} \underbrace{\left(1+\frac{x}{a}\right)^{n}}_{\begin{array}{c} \text { apply binomial } \\ \text { to this } \end{array}} .

Using Eq. (2.12) and, assuming $s$ is small, we approximate $\sqrt{1+s}$ by

(1+s)^{1 / 2}=1+\frac{s}{2}-\frac{s^{2}}{8}+\cdots

using the expansion above in the limit $s \rightarrow 0$ , we observe that the limit (2.9) approaches the value of $1 / 2$ , as shown above.

Example 2.3 Evaluate the limit,

\lim _{x \rightarrow \infty} x\left(\sqrt{1+x^{2}}-\sqrt{x^{2}-1}\right) .

Solution The limit is of the form $\infty-\infty$ . Again, we have more than one option here; we observe that we can simplify the function and evaluate the limit by multiplying and dividing by $\left(\sqrt{1+x^{2}}+\sqrt{x^{2}-1}\right)$ such that

\begin{aligned} \lim _{x \rightarrow \infty} x\left(\sqrt{1+x^{2}}-\sqrt{x^{2}-1}\right) & =\lim _{x \rightarrow \infty} x\left(\sqrt{1+x^{2}}-\sqrt{x^{2}-1}\right) \frac{\left(\sqrt{1+x^{2}}+\sqrt{x^{2}-1}\right)}{\left(\sqrt{1+x^{2}}+\sqrt{x^{2}-1}\right)} \\ & =\frac{2 x}{\left(\sqrt{1+x^{2}}+\sqrt{x^{2}-1}\right)} \end{aligned}

where, in the limit $x \rightarrow \infty$ , the denominator becomes $2 x$ which implies that the limit (2.13) is 1 .

Alternatively, we can use the binomial series (2.12). First, introducing a change of variable $x=1 / s$ such that (2.13) becomes

\lim _{s \rightarrow 0} \frac{1}{s}\left(\sqrt{1+\frac{1}{s^{2}}}-\sqrt{\frac{1}{s^{2}}-1}\right)=\lim _{s \rightarrow 0} \frac{1}{s^{2}}\left[\left(1+s^{2}\right)^{1 / 2}-\left(1-s^{2}\right)^{1 / 2}\right] .

For small $s$ , applying the binomial expansion (2.12) with $x= \pm s^{2}$ gives the limit as above.

We have already seen the use of the binomial series in evaluating certain limits. We move on to geometric series. The sum of a geometric series can help us compute limits like,

\lim _{x \rightarrow 1} \frac{x^{n}-1}{x-1} .

Definition 2.2 A geometric series is any series that can be written in the form

\sum_{n=1}^{\infty} a r^{n-1}

or, starting at $n=0$ , we can equivalently write,

\sum_{n=0}^{\infty} a r^{n}

where $r$ is referred to as the common ratio. From (2.18), a finite geometric series has the form

\sum_{m=0}^{n} a x^{m}=a+a x+a x^{2}+a x^{3}+\cdots+a x^{n} .

It is possible to find the sum of a finite geometric series. ${ }^{4}$ To find the sum, multiply $(2.19)$ by $x$ ,

x \sum_{m=0}^{n} a x^{m}=a x+a x^{2}+a x^{3}+\cdots+a x^{n+1} .

Subtracting (2.20) from (2.19) yields

\sum_{m=0}^{n} a x^{m}=\frac{a\left(x^{n+1}-1\right)}{x-1} .

Equivalently, we have

\sum_{m=1}^{n} a x^{m-1}=\frac{a\left(x^{n}-1\right)}{x-1} .

We use this sum in Example 2.4 to evaluate the limit in Eq. (2.16).

Example 2.4 Evaluate the limit given in Eq. (2.16).

Solution We choose to use Eq. (2.22) here but Eq. (2.21) can also be used. For $a=1$ in Eq. (2.22), we have

\frac{x^{n}-1}{x-1}=\sum_{m=1}^{n} x^{m-1}

For $x \neq 1$ , the RHS of Eq. (2.23) gives

\sum_{m=1}^{n} x^{m-1}=1+\underbrace{x+x^{2}+x^{3}+\cdots x^{n-1}}_{n-1 \text { terms }},

and as $x \rightarrow 1$ the sum (and therefore the limit) is $n$ .

Trigonometric limits

Next, we consider the limit

\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta},

which is of the form 0/0. L'Hôpital's rule (see Section 2.3) shows that the limit is 1 . We can also show that the limit is 1 geometrically. Consider a sector of a circle of radius $r$ embedded in a right triangle as shown below (the red, dotted line shows part of the circle).

${ }^{4}$ We can also find the sum of an infinite geometric series provided that $|r|<1$ .

We can see from the diagram that

(area of triangle $\mathrm{OAB})<($ area of sector $\mathrm{OAB})<($ area of triangle OBC $)$ .

The area of triangle OBC is $r \overline{B C}$ but $\overline{B C}=r \tan \theta$ and so

\text { area of triangle } \mathrm{OBC}=r^{2} \tan \theta

Next, the area of sector $O A B$ is $(\theta / 2 \pi) \pi r^{2}$ which simplifies to

\text { area of sector } \mathrm{OAB}=\frac{1}{2} r^{2} \theta \text {. }

Finally, the triangle $\mathrm{OAB}$ is an isosceles triangle since $\overline{O B}=\overline{O A}=r$ . The grey, dotted line in the diagram splits the isosceles triangle into two right triangles each with a 'base' of length $\overline{A B} / 2$ . Simple trigonometry shows that the height of the triangle (i.e. the length of the grey, dotted line) is $h=r \cos (\theta / 2)$ and $\overline{A B} / 2=r \sin (\theta / 2)$ . Since the area of the triangle is given by $(1 / 2) \times$ height $\times$ base,

\text { area of triangle } \mathrm{OAB}=\frac{1}{2} r^{2} \sin \theta ;

where we have made use of the trigonometric identity $\sin \theta \equiv 2 \sin (\theta / 2) \cos (\theta / 2)$ . Going back to (2.26), we have

\frac{1}{2} r^{2} \sin \theta<\frac{1}{2} r^{2} \theta<\frac{1}{2} r^{2} \tan \theta

which simplifies to

\sin \theta<\theta<\tan \theta

Taking reciprocals gives

\frac{1}{\sin \theta}>\frac{1}{\theta}>\cot \theta

and multiplying through by $\sin \theta$ (which is positive in $0<\theta<\pi$ ) gives

1>\frac{\sin \theta}{\theta}>\cos \theta

Now, $\sin \theta / \theta$ is sandwiched between $\cos \theta$ and 1 . We observe that $\cos \theta \rightarrow 1$ as $\theta \rightarrow 0$ , then $\sin \theta / \theta$ as $\theta \rightarrow 0$ must also approach 1; this proves that the value of the limit in Eq. (2.25) is 1 . Using the limit in (2.25), we can obtain other limits; some examples are given next.

Consider

\lim _{x \rightarrow 0} \frac{\tan x}{x}=1

since

\frac{\tan x}{x}=\frac{\sin x}{x} \cdot \frac{1}{\cos x} \rightarrow 1 \cdot 1=1

as $x \rightarrow 0$ .

The limit

\lim _{x \rightarrow \pi / 2} \frac{\cos x}{x-\pi / 2}

is of the indeterminate form $0 / 0$ . We can rewrite as

\lim _{x \rightarrow \pi / 2} \frac{\cos x}{x-\pi / 2}=\lim _{x \rightarrow \pi / 2}-\frac{\sin (x-\pi / 2)}{x-\pi / 2},

where we have made use of the identity given by Eq. (1.7). Further, letting $s=x-\pi / 2$ ,

\lim _{x \rightarrow \pi / 2}-\frac{\sin (x-\pi / 2)}{x-\pi / 2}=\lim _{s \rightarrow 0}-\frac{\sin s}{s} \rightarrow-1 .

Lastly, another important limit is

\lim _{n \rightarrow \infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}

This limit is given here without proof. The proof can be obtained using power series (see Chapter 6) or, alternatively, using first order ordinary differential equations (ODEs) whose solutions are discussed in Chapter $12^{5}$ .

Limits: Several Variables L'Hôpital's rule