How to Calculate Limits III

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*}

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