In this posting, we discuss limits of trigonometric functions. The most basic trigonometric functions are of course \(y=\sin x\) and \(y=\cos x\). They have the following limit properties.

Theorem 5. For any \(a\in\mathbb R\), \[\lim_{x\to a}\sin x=\sin a,\ \lim_{x\to a}\cos x=\cos a.\]

You notice that both \(y=\sin x\) and \(y=\cos x\) satisfy the same limit property as polynomial functions (Theorem 2 in Lecture 4). This is not a coincidence and the reason behind this is that polynomial functions, \(y=\sin x\) and \(y=\cos x\) are continuous functions. This will become clear when we discuss the continuity of a function later. Limit properties of other trigonometric function will stem out automatically from the above Theorem 5 and Theorem 1 in Lecture 4. For example, the limit property of \(y=\tan x\) is given by \[\lim_{x\to a}\tan x=\lim_{x\to a}\frac{\sin x}{\cos x}=\frac{\sin a}{\cos a}=\tan a,\] where \(\tan a\) is defined or equivalently \(\cos a\ne 0\).

Theorem 6. Suppose that \(f(x)\leq g(x)\) near \(x=a\) and both \(\displaystyle\lim_{x\to a}f(x)\), \(\displaystyle\lim_{x\to a}g(x)\) exist. Then \[\lim_{x\to a}f(x)\leq \lim_{x\to a}g(x).\]

Corollary 7. [Squeeze Theorem, Sandwich Theorem] Suppose that \(f(x)\leq g(x)\leq h(x)\) near \(x=a\). If \(\displaystyle\lim_{x\to a}f(x)=\lim_{x\to a}h(x)=L\) then \[\lim_{x\to a}g(x)=L.\]

Squeeze Theorem is useful to calculate certain type of limits such as the following example.

Example. Find the limit \(\displaystyle\lim_{x\to 0}x^2\sin\frac{1}{x}\).

Solution. Since \(-1\leq\sin\frac{1}{x}\leq 1\), \[-x^2\leq x^2\sin\frac{1}{x}\leq x^2\] for all \(x\ne 0\). Since \(\displaystyle\lim_{x\to 0}(-x^2)=\lim_{x\to 0}x^2=0\), by Squeeze Theorem \[\lim_{x\to 0}x^2\sin\frac{1}{x}=0.\] The following picture also confirms our result.

There is another important limit that involves a trigonometric function.  It is

Theorem 8. \(\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1\).

This is an important formula. You will readily see that this limit is \(\frac{0}{0}\) type indeterminate form. So this means that \(\sin x\) must have a factor  \(x\) in it. But how do we factor \(\sin x\)? It is not a polynimial! In fact. it is (sort of). This is something you are going to learn in Calculus 3 (MAT 169) but I want you to taste it. The function \(\sin x\) is can be written as a never-ending polynomial (such a polynomial is called a power series in mathematics) \[\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots,\]where \(n!\) denotes the \(n\) factorial\[n!=n(n-1)(n-2)(n-3)\cdots3\cdot 2\cdot 1.\] So \begin{eqnarray*}\lim_{x\to 0}\frac{\sin x}{x}&=&\lim_{x\to 0}\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots}{x}\\&=&\lim_{x\to 0}\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots\right)\\&=&1.\end{eqnarray*}

We have now confirmed that the formula is indeed correct, but is there a more fundamental proof without using power series? Yes, there is. In fact it can be proved using trigonometry. First consider the case when \(x\to 0+\). In this case, without loss of generality we may assume that \(x\) is an acute angle so we have the following picture.

The areas of \(\triangle OAC\), \(\sphericalangle OBC\) and \(\triangle OBD\) are, respectively, given by \(\frac{1}{2}\cos x\sin x\), \(\frac{1}{2}x\) and \(\frac{1}{2}\tan x\). Clearly from the picture they satisfy the inequality \[\frac{1}{2}\cos x\sin x<\frac{1}{2}x<\frac{1}{2}\tan x.\] Dividing this inequality by \(\frac{1}{2}\sin x\) (note that \(\sin x>0\) since \(x\) is an acute angle) we obtain\[\cos x<\frac{x}{\sin x}<\frac{1}{\cos x}\] or equivalently,\[\frac{1}{\cos x}<\frac{\sin x}{x}<\cos x.\] Now \(\displaystyle\lim_{x\to 0+}\cos x=\lim_{x\to 0+}\frac{1}{\cos x}=1\), so by Squeeze Theorem,\[\lim_{x\to 0+}\frac{\sin x}{x}=1.\] Similarly, we can also show that\[\lim_{x\to 0-}\frac{\sin x}{x}=1.\] Hence completes the proof.

Example. Find $\displaystyle\lim_{x\to 0}\frac{\sin 7x}{4x}$.

Solution. \begin{eqnarray*}\lim_{x\to 0}\frac{\sin 7x}{4x}&=&\lim_{x\to 0}\frac{7}{4}\frac{\sin 7x}{7x}\\&=&\frac{7}{4}\lim_{x\to 0}\frac{\sin 7x}{7x}\\&=&\frac{7}{4}\ \left(\lim_{x\to 0}\frac{\sin 7x}{7x}=1\right).\end{eqnarray*}

Example. Find $\displaystyle\lim_{\theta\to 0}\frac{\cos\theta-1}{\theta}$.

Solution. \begin{eqnarray*}\lim_{\theta\to 0}\frac{\cos\theta-1}{\theta}&=&\lim_{\theta\to 0}\frac{\cos\theta-1}{\theta}\frac{\cos\theta+1}{\cos\theta+1}\\&=&\lim_{\theta\to 0}\frac{\cos^2\theta-1}{\theta(\cos\theta+1)}\\&=&\lim_{\theta\to 0}\frac{\cos^2\theta-1}{\theta(\cos\theta+1)}\\&=&\lim_{\theta\to 0}\frac{-\sin^2\theta}{\theta(\cos\theta+1)}\\&=&-\lim_{\theta\to 0}\frac{\sin\theta}{\theta}\frac{\sin\theta}{\cos\theta+1}\\&=&-\lim_{\theta\to 0}\frac{\sin\theta}{\theta}\cdot\lim_{\theta\to 0}\frac{\sin\theta}{\cos\theta+1}\\&=&-1\cdot 0=0.\end{eqnarray*}

There may be a confusion regarding the meaning of “limit undefined”. It is actually a matter of opinion. My notion of “limit undefined” is different from that of your textbook. In your textbook, the limit is said to be undefined if it fails to exist as a number. So for instance the limit \[\lim_{x\to 0}\frac{1}{x^2}=\infty\] is undefined according to textbook. In my case however I would still say that the limit exists as \(\infty\) since the left-hand limit and the right-hand limit coincide as \(\infty\). It wouldn’t matter whichever definition you follow as long as you are clear about it.

In the previous posting, we studied how to calculate limit of a rational function (Corollary 3). Let us state it here again:

Corollary 3. [Limit of a Rational Function] Let \(p(x)\) and \(q(x)\) be two polynomials. Then for any real number \(b\), \[\lim_{x\to b}\frac{p(x)}{q(x)}=\frac{p(b)}{q(b)}\] provided \(q(b)\ne 0\).

But what if \(q(b)=0\)? To answer this question let us take a look at the following example.

Example. Find the limit \(\displaystyle\lim_{x\to -1}\frac{x^2+3x+2}{x^2-x-2}\).

Solution. Let \(p(x)=x^2+3x+2\) and \(q(x)=x^2-x-2\). Then \(p(-1)=0\) and \(q(-1)=0\). Since \(q(-1)=0\), we cannot use Corollary 3 to calculate the limit. So what do we do? Note that \(p(-1)=0\) and \(q(-1)=0\) means that both \(p(x)\) and \(q(x)\) contains a power of \((x+1)\) in them. Let us factor out the maximum common power of \((x+1)\) from \(p(x)\) and \(q(x)\). Since \(x\to -1\), \(x\ne -1\) i.e. \(x+1\ne 0\). So we can cancel the maximum common power of \((x+1)\) and then calculate limit of the resulting function as \(x\to -1\): \begin{eqnarray*}\lim_{x\to -1}\frac{x^2+3x+2}{x^2-x-2}&=&\lim_{x\to -1}\frac{(x+1)(x+2)}{(x-2)(x+1)}\\&=&\lim_{x\to -1}\frac{x+2}{x-2}\ \mbox{since \(x\ne -1\)}\\&=&-\frac{1}{3}.\end{eqnarray*}

Remark. [Indeterminate Form] In the above example, \[\frac{\displaystyle\lim_{x\to -1}(x^2+3x+2)}{\displaystyle\lim_{x\to -1}(x^2-x-2)}=\frac{0}{0}.\] What is this? and how do we understand it? It turns out that the quantity \(\frac{0}{0}\) is not undefined but something else. Remember that here \(0\) is not a number but an infinitesimal, a state that is extremely close to the number \(0\). The quantity \(\frac{0}{0}\) is called an indeterminate form. There are other types of indeterminate forms, to name a few, \(\frac{\infty}{\infty}\), \(0\cdot\infty\), \(0^0\), etc. We will study them later. There are four possibilities for the value of an indeterminate form: \(0\), \(\pm\infty\), or a non-zero real number. Although we denote infinitesimals by the same symbol \(0\), some infinitesimals dominate others. For instance, consider the limit of a rational function \(\displaystyle\lim_{x\to a}\frac{p(x)}{q(x)}\). Suppose that \(\displaystyle\lim_{x\to a}p(x)=\lim_{x\to a}q(x)=0\). There can be three possible scenarios then:

1. If \(p(x)\) approaches \(0\) way faster than \(q(x)\) does, then \(\displaystyle\lim_{x\to a}\frac{p(x)}{q(x)}=0\).
2. If \(q(x)\) approaches \(0\) way faster than \(p(x)\) does, then \(\displaystyle\lim_{x\to a}\frac{p(x)}{q(x)}=\pm\infty\).
3. If \(p(x)\) and \(q(x)\) approaches \(0\) at about the same rate (speed), then \(\displaystyle\lim_{x\to a}\frac{p(x)}{q(x)}\) may be a non-zero real number.

Example. Find the limit \[\lim_{x\to 2}\frac{4-x^2}{3-\sqrt{x^2+5}}.\]

Solution. \(\displaystyle\lim_{x\to 2}(4-x^2)=\lim_{x\to 2}(3-\sqrt{x^2+5})=0\). This means that both the numerator and the denominator have a power of \(x-2\) as a common factor. As we did in the previous example, we would attempt to factor both the numerator and the denominator. Only problem is that the denominator is not a polynomial and we don’t know how to factor it. Well, we learned about rationalizing the denominator in algebra. We multiply the numerator and the denominator by the conjugate \(3+\sqrt{x^2+5}\) of the denominator. More specifically,\begin{eqnarray*}\lim_{x\to 2}\frac{4-x^2}{3-\sqrt{x^2+5}}&=&\lim_{x\to 2}\frac{4-x^2}{3-\sqrt{x^2+5}}\cdot\frac{3+\sqrt{x^2+5}}{3+\sqrt{x^2+5}}\\&=&\lim_{x\to 2}\frac{(4-x^2)(3+\sqrt{x^2+5})}{4-x^2}\\&=&\lim_{x\to 2}(3+\sqrt{x^2+5})\\&=&6.\end{eqnarray*}

When you calculate limits, the following theorem plays a crucial role.

Theorem 1. Suppose that \(c\) is a constant and the limits \[\lim_{x\to a}f(x)\ {\rm and}\ \lim_{x\to a}g(x)\] exist. Then the following properties hold:

1. \(\displaystyle\lim_{x\to a}\{f(x)+g(x)\}=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)\)
2. \(\displaystyle\lim_{x\to a}cf(x)=c\lim_{x\to a}f(x)\)
3. \(\displaystyle\lim_{x\to a}f(x)g(x)=\lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x)\)
4. \(\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\displaystyle\lim_{x\to a}f(x)}{\displaystyle\lim_{x\to a}g(x)}\) provided \(\displaystyle\lim_{x\to a}g(x)\ne 0\)

Before we discuss how we can calculate limits, there are very basic limits we need to know. They are: \[\lim_{x\to a}x=a\ {\rm and}\ \lim_{x\to a}c=c,\] where \(c\) is a constant. The limits are trivial from our intuition and also from their graphs. However, those who have analytical mind may want prove them. They can be proved using the Cauchy’s definition of a limit we discussed in the previous posting. Let us first prove \(\displaystyle\lim_{x\to a}x=a\).

Proof. Let \(\epsilon>0\) be given. Choose \(\delta=\epsilon\). Then for all \(x\) that satisfies the inequality \[|x-a|<\delta,\] it is true that \[|x-a|<\epsilon.\]

Now we prove \(\displaystyle\lim_{x\to a}c=c,\) where \(c\) is a constant.

Proof. Let \(\epsilon>0\) be given. Choose \(\delta>0\) to be any positive real number. Then for all \(x\) that satisfies the inequality \[|x-a|<\delta,\] we have \[|c-c|=0<\epsilon.\]

Using these two limits, we can now show the following useful theorem for calculating limits of polynomial functions.

Theorem 2. Let \(p(x)\) be a polynomial. Then for any real number \(b\), \[\lim_{x\to b}p(x)=p(b).\]

Proof. Let \(p(x)\) be a polynomial of degree \(n\). Then \(p(x)\) can be written as \[p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0,\] where \(a_n,a_{n-1},\cdots,a_2,a_1,a_0\) are constant real coefficients. Now \begin{eqnarray}\lim_{x\to b}p(x)&=&\lim_{x\to b}(a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0)\\&=&\lim_{x\to b}a_nx^n+\lim_{x\to a}a_{n-1}x^{n-1}+\cdots+\lim_{x\to a}a_2x^2+\lim_{x\to a}a_1x+\lim_{x\to a}a_0,\ \mbox{by property 1}\\&=&a_n\lim_{x\to b}x^n+a_{n-1}\lim_{x\to b}x^{n-1}+\cdots+a_2\lim_{x\to b}x^2+a_1\lim_{x\to b}x+a_0,\ \mbox{by property 2 and \(\lim_{x\to b}a_0=a_0\)}\\&=&a_nb^n+a_{n-1}b^{n-1}+\cdots+a_2b^2+a_1b+a_0,\ \mbox{by \(\lim_{x\to b}x=b\) and property 3}\\&=&p(b).\end{eqnarray}

Theorem 2 also implies that any polynomial function is continuous everywhere. We will discuss the notion of continuity later.

Due to theorem 2 and property 4 of theorem 1, the following corollary about the limit of a rational function holds.

Corollary 3. [Limit of a Rational Function] Let \(p(x)\) and \(q(x)\) be two polynomials. Then for any real number \(b\), \[\lim_{x\to b}\frac{p(x)}{q(x)}=\frac{p(b)}{q(b)}\] provided \(q(b)\ne 0\).

Theorem 4. [Other Important Limits] Suppose that \(\displaystyle\lim_{x\to a}f(x)\) exists. Then the following properties hold:

1. \(\displaystyle\lim_{x\to a}f(x)^n=[\displaystyle\lim_{x\to a}f(x)]^n\)
2. \(\displaystyle\lim_{x\to a}\root n\of{f(x)}=\root n\of{\displaystyle\lim_{x\to a}f(x)}\)
3. \(\displaystyle\lim_{x\to a}\ln f(x)=\ln[\displaystyle\lim_{x\to a}f(x)]\)
4. \(\displaystyle\lim_{x\to a}\sin f(x)=\sin[\displaystyle\lim_{x\to a}f(x)]\)
5. \(\displaystyle\lim_{x\to a}\cos f(x)=\cos[\displaystyle\lim_{x\to a}f(x)]\)

It is assumed that \(\displaystyle\lim_{x\to a} f(x)\) belongs to the domain of each function in each property. For example, in property 2 if \(n\) is an even integer, it is required that \(\displaystyle\lim_{x\to a} f(x)\geq 0\). The first property is a direct consequence of property 3 of theorem 1. The rest of the properties are related to the continuity of a function. This will be discussed later.