I have the following function:
$$\lim_{x\to1} {\sin(\pi x)\over{1-x}}.$$
I need to calculate the limit, although I can't use here L'Hôpital's rule.
I have a clue that says to use a correct mathematical placement.
Can you please advise only what the placement should be?
I've tried several placements but couldn't find the correct one.
Thank you all.
Best Answer
We want to compute the limit
$$\lim_{x\to 1}\frac{\sin(\pi x)}{x-1}.$$
Notice first that, using the sine sum formula, we have that
$$\sin(\pi x)=\sin(\pi x-\pi+\pi)=\sin(\pi x-\pi)\underbrace{\cos(\pi)}_{=-1}+\cos(\pi x-\pi)\underbrace{\sin(\pi)}_{=0}=-\sin(\pi x-\pi).$$
We can thus rewrite
$$\frac{\sin(\pi x)}{x-1}=-\pi\frac{\sin(\pi x-\pi)}{\pi x-\pi}.$$
Now as $\pi x-\pi\to 0$ as $x\to 1$, we can use the standard limit
$$\lim_{t\to 0}\frac{\sin(t)}{t}=1$$
to get that
$$\lim_{x\to 1}\frac{\sin(\pi x)}{x-1}=-\pi\left(\lim_{x\to 1}\frac{\sin(\pi x-\pi)}{\pi x-\pi}\right)=-\pi.$$