Trigonometric Limit Question

Question

I am beginning trigonometric limits. I believe this limit requires a substitution but I am not quite sure exactly how the substitution works. Any explanations on the process of this specific question would be wonderful. Given:

$$\lim_{x \to \pi} \dfrac{\sin x}{\pi-x}$$

I need to determine the limit. I think I have to rewrite the limit approaching part and I am guessing this will involve the following:

$$\lim_{x \to 0} \dfrac{\sin x}{x}=1$$

Note: my course so far has not covered L'hospitals rule.

Thanks

There are many good answers provided. To be more careful, the substitution trick works because $f(x)=\pi-x$ is a continuous function. — , Oct 07 '14 at 02:54

score 7 · Accepted Answer · answered Oct 07 '14 at 02:50

7

Let $t=\pi-x$. Then as $x\to \pi$, $t\to 0$.

So $$\lim_{x \to \pi} \dfrac{\sin x}{\pi-x}=\lim_{t \to 0} \dfrac{\sin (\pi-t)}{t}=\lim_{t \to 0} \dfrac{\sin t}{t}=1.$$

answered Oct 07 '14 at 02:50

Paul

20,553

Very clear, thank you! – nitrous2 Oct 07 '14 at 03:02
@nitrous2: You are welcome! – Paul Oct 07 '14 at 03:07

E W H Lee · Answer 2 · 2014-10-07T03:01:05.503

2

Since $\sin x = -\sin(x - \pi)$, $$ \lim_{x \rightarrow \pi} \frac{\sin x}{\pi-x} = \lim_{x \rightarrow \pi} \frac{\sin (x-\pi)}{x-\pi} = 1. $$ Alternately, $$ \lim_{x \rightarrow \pi} \frac{\sin x}{\pi-x} = - \lim_{x \rightarrow \pi} \frac{\sin x - \sin \pi}{x-\pi} = - \frac{d}{dx}\sin x \Bigg|_{x=\pi} = 1. $$

edited Oct 07 '14 at 03:01

answered Oct 07 '14 at 02:50

E W H Lee

2,336

Trigonometric Limit Question

2 Answers2