In this lecture note, 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. 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 as seen here. 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 functions stem out of the above theorem along with limit laws discussed here. 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. 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. [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 for calculating a 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.
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| The graph of $y=x^2\sin\frac{1}{x}$ |
There is another important limit that involves a trigonometric function. It is
Theorem. \(\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1\).
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 polynomial! In fact. it is sort of. The function \(\sin x\) is can be written as a never-ending polynomial (such a polynomial is called a power series) \[\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots,\]where \(n!\) denotes the \(n\) factorial\[n!=n(n-1)(n-2)(n-3)\cdots3\cdot 2\cdot 1.\] So \begin{align*}\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{align*}
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.
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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 this completes the proof.
Example. Find $\displaystyle\lim_{x\to 0}\frac{\sin 7x}{4x}$.
Solution. \begin{align*}\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{align*}
Example. Find $\displaystyle\lim_{\theta\to 0}\frac{\cos\theta-1}{\theta}$.
Solution. \begin{align*}\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{align*}
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