The limit of the function $(1-x^2)/\sin(\pi x)$ as $x\to 1$ What is $${\lim_{x \to 1}} \frac{1-x^2}{\sin(\pi x)} \text{ ?} $$ I got it as $0$ but answer in the book as $2/ \pi$. Can you guys tell me what's wrong?
 A: Let $x-1=t$, then
\begin{align}
\lim_{x\to 1}\frac{1-x^2}{\sin \pi x} &= \lim_{t\to 0}\frac{-t(t+2)}{\sin (\pi(t+1))}\\
&=\lim_{t\to 0} \left(\frac{t}{\sin \pi t}\cdot (t+2)\right)\\
&=\frac{2}{\pi}
\end{align}
A: Seems that using the L'Hospital rule we trivially get the answer in the book.
Differentiating the numerator and denominator we get:   
$$\frac{-2x}{\cos(\pi x) \cdot \pi} $$
Since the above function is continuous at $x=1$, the limit is:   
$$\frac{-2.1}{\cos(\pi \cdot 1) \cdot \pi} $$
$$\frac{-2\cdot1}{-1 \cdot \pi} = \frac{2}{\pi}$$
A: You have 
 $${1-x^2 \over \sin(\pi x)}= {(1-x)(1+x)\over \sin(\pi x)}\sim {2(1-x)\over \sin(\pi x)}. $$
Now look at the definition of the derivative for $\sin$ at $\pi$.
A: \begin{align}
& \lim_{x\to1} \frac{(1-x)(1+x)}{\sin(\pi x)} = \lim_{x\to1} \frac{(1+x)(1-x)}{\sin(\pi(1-x))} \\[10pt]
= {} & \underbrace{\left( \lim_{x\to1} (1+x) \right)  \left( \frac 1 \pi \lim_{x\to1} \frac{\pi(1-x)}{\sin(\pi(1-x)} \right)}_{\text{(See the remark on this below.)}} = 2 \cdot \frac 1 \pi \lim_{u\to 0} \frac u {\sin u}
\end{align}
where $u=\pi(1-x)$.  That the last limit is $1$ is something you would be presumed to have seen before.
Breaking it into two limits like that is permissible if both limits exist and are finite numbers, and in this case we see that they are.
