I know similar questions have been asked but I could not find any answer that applies to my case, so I'll ask anyway. I'm having trouble getting rid of an indeterminate without using derivatives
$$\lim_{x\to 1} \frac{\sin(\pi x)}{x-1}$$
One idea I had was to multiply both sides of the fractions by $$\pi x$$ like this $$\lim_{x\to 1} \frac{\sin(\pi x)\pi x}{\pi x^2-\pi x}$$ to cancel out the sine but then I realized that I could not do that because x tends to 1, and not to 0. What can I do?
On
With equivalents:
Set $x=1+h$. We know $\sin u\sim_0u$, so $$\frac{\sin\pi x}{x-1}=\frac{\sin(\pi+\pi h)}{h}=-\frac{\sin\pi h}h\sim_0-\frac{\pi h}h=-\pi.$$
On
$$ \sin \pi x = \sin (\pi(x-1) + \pi) = - \sin (\pi(x-1)) $$ SO $$ \lim_{x \to 1} \frac{\sin \pi x}{x-1}= -\lim_{x \to 1}\frac{\sin \pi(x-1)}{x-1} = -\pi $$
On
\begin{eqnarray} \lim_{x \rightarrow 1}\frac{\sin(\pi x)}{1-x} &=&\lim_{h \rightarrow 0}\frac{\sin(\pi (1-h))}{h} \\ &=&\lim_{h \rightarrow 0}\frac{\sin(-\pi h)}{h} \\ &=&\lim_{h \rightarrow 0}\frac{-\sin(\pi h)\pi}{\pi h} \tag{common limit} \\ &=&-\pi. \end{eqnarray}
On
Substitute $\pi x=t+\pi$, so $x-1=\frac{t}{\pi}$; for $x=1$ we have $t=0$, so the limit becomes $$ \lim_{x\to1}\frac{\sin(\pi x)}{x-1}= \lim_{t\to0}\frac{\sin(t+\pi)}{t/\pi}= \lim_{t\to0}-\pi\frac{\sin t}{t} $$
On the other hand the limit is the derivative at $1$ of the function $f(x)=\sin(\pi x)$, and $$ f'(x)=\pi\cos(\pi x) $$ so $$ f'(1)=-\pi $$
(Note: this is not l’Hôpital.)
$$\lim_{h\rightarrow0}\frac{\sin(\pi(1-h))}{1-1-h}$$ $$\lim_{h\rightarrow0}\frac{\sin(\pi h)\pi}{-\pi h}\tag{Multiply and divide by pi}$$ $$-\pi$$