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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

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    $\begingroup$ You can use the fact that $\sin x = -\sin(x-\pi)$. $\endgroup$ – E W H Lee Oct 7 '14 at 2:47
  • $\begingroup$ But how does the limit aspect work? The substitution part? $\endgroup$ – nitrous2 Oct 7 '14 at 2:49
  • $\begingroup$ There are many good answers provided. To be more careful, the substitution trick works because $f(x)=\pi-x$ is a continuous function. $\endgroup$ – user175968 Oct 7 '14 at 2:54
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Let $t=\pi-x$. Then as $x\to \pi$, $t\to 0$.

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

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  • $\begingroup$ Very clear, thank you! $\endgroup$ – nitrous2 Oct 7 '14 at 3:02
  • $\begingroup$ @nitrous2: You are welcome! $\endgroup$ – Paul Oct 7 '14 at 3:07
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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, $$ \lim_{x \rightarrow \pi} \frac{\sin x}{\pi-x} = - \lim_{x \rightarrow \pi} \frac{\sin x - \sin \pi}{x-\pi} = - \frac{d}{dx}\sin x \Bigg|_{x=\pi} = 1. $$

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