Trigonometric Limit 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

• You can use the fact that $\sin x = -\sin(x-\pi)$. – E W H Lee Oct 7 '14 at 2:47
• But how does the limit aspect work? The substitution part? – nitrous2 Oct 7 '14 at 2:49
• There are many good answers provided. To be more careful, the substitution trick works because $f(x)=\pi-x$ is a continuous function. – user175968 Oct 7 '14 at 2:54

Let $t=\pi-x$. Then as $x\to \pi$, $t\to 0$.
So $$\lim_{x \to \pi} \dfrac{\sin x}{\pi-x}=\lim_{t \to 0} \dfrac{\sin (\pi-t)}{t}=\lim_{t \to 0} \dfrac{\sin t}{t}=1.$$
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, $$\lim_{x \rightarrow \pi} \frac{\sin x}{\pi-x} = - \lim_{x \rightarrow \pi} \frac{\sin x - \sin \pi}{x-\pi} = - \frac{d}{dx}\sin x \Bigg|_{x=\pi} = 1.$$