Limit of trigonometrical function $\sin(\pi x)/{(1-x)}$ as $x\to1$?

Question

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.

Answer 1

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

$$\lim_{x\to 1}\frac{\sin(\pi x)}{x-1}=-\pi\left(\lim_{x\to 1}\frac{\sin(\pi x-\pi)}{\pi x-\pi}\right)=-\pi.$$

Answer 2

You write $$\frac{\sin \pi x}{1-x}=-\pi\times\frac{\sin \pi (x-1)}{\pi(x-1)}$$ and you know that the $\lim _{t\to 0}\frac{\sin t}{t}=1$. Hence $\lim \limits_{x\to 1}\frac{\sin \pi (x-1)}{\pi(x-1)}=1.$ So the answer should be $-\pi$.

Answer 3

You could use $$\pi x = \pi x + \pi - \pi = \pi(x-1) + \pi $$

and with $t = \pi(x-1)$ if $x\to 1$ then $t \to 0$