<s>[INST] Evaluate $\lim_{x \to 1}\frac{1-x^2}{\sin\pi x}$ without L'Hopital I'm trying to evaluate the given limit without using L'hopitals rule. $$\lim_{x \to 1}\frac{1-x^2}{\sin(\pi x)}$$
Replacing $x$ by $1$ leads to $\frac{0}{0}$.
I have tried multiplying by $\frac{1+x^2}{1+x^2}$ and resoving the $\sin x$ factor by doing:
$$\frac{x}{\sin x} \cdot \frac{(\frac{1}{x} -x)}{\pi}$$
so $\frac{x}{\sin x} = 1$ but then the result still is $\frac{0}{\pi}$. All other options I see also lead to $\frac{0}{0}$ and I'm too sure how to proceed. Any hints are very welcome.
 [/INST] Write
$$\frac{1-x^2}{\sin\pi x}=\frac{1-x^2}{\sin(\pi-\pi x)}=\frac{1-x^2}{\sin\pi(1- x)}=\frac{(1-x)(1+x)}{2\sin\frac{\pi(1- x)}{2}\cos\frac{\pi(1- x)}{2}}=\frac12\times\frac{2}{\pi}\frac{\frac{\pi(1- x)}{2}}{\sin\frac{\pi(1- x)}{2}}\times\frac{1+x}{\cos\frac{\pi(1- x)}{2}}$$
so $t=1-x$ we have
$$\lim_{x\to1}\frac{1-x^2}{\sin\pi x}=\lim_{t\to0}\frac12\times\frac{2}{\pi}\frac{\frac{\pi t}{2}}{\sin\frac{\pi t}{2}}\times\frac{2-t}{\cos\frac{\pi t}{2}}=\frac12\times\frac{2}{\pi}\times\frac{2}{1}=\color{red}{\frac{2}{\pi}}$$
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