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\mathrm{\text { Evaluate } \quad \lim _{x \rightarrow 1} \frac{\sin (\pi x)}{1-x}}

Option: 1

-\pi


Option: 2

0


Option: 3

\pi


Option: 4

1


Answers (1)

best_answer

We want to compute the limit

                                            \mathrm{ \lim _{x \rightarrow 1} \frac{\sin (\pi x)}{x-1} . }

Notice first that, using the sine sum formula, we have that

\mathrm{ \sin (\pi x)=\sin (\pi x-\pi+\pi)=\sin (\pi x-\pi) \underbrace{\cos (\pi)}_{=-1}+\cos (\pi x-\pi) \underbrace{\sin (\pi)}_{=0}=-\sin (\pi x-\pi) . }
We can thus rewrite

                                        \mathrm{ \frac{\sin (\pi x)}{x-1}=-\pi \frac{\sin (\pi x-\pi)}{\pi x-\pi} . }

Now as \mathrm{\pi x-\pi \rightarrow 0} as \mathrm{x \rightarrow 1}, we can use the standard limit

                                            \mathrm{ \lim _{t \rightarrow 0} \frac{\sin (t)}{t}=1 }
to get that

                              \mathrm{ \lim _{x \rightarrow 1} \frac{\sin (\pi x)}{x-1}=-\pi\left(\lim _{x \rightarrow 1} \frac{\sin (\pi x-\pi)}{\pi x-\pi}\right)=-\pi . }

Posted by

Ajit Kumar Dubey

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