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.
On
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 $$\lim_{x\to1}\frac{1-x^2}{\sin\pi x}=\lim_{t\to0}\frac12\times\frac{2}{\pi}\frac{\frac{\pi t}{2}}{\sin\frac{\pi t}{2}}\times\frac{2-t}{\cos\frac{\pi t}{2}}=\frac12\times\frac{2}{\pi}\times\frac{2}{1}=\color{red}{\frac{2}{\pi}}$$
Using the substitution $x=1-y$ and the property $\sin(\pi-z)=\sin(z)$ we have
$$ \lim_{x\to 1}\frac{(1-x)(1+x)}{\sin(\pi x)}=\lim_{y\to 0}\frac{y(2-y)}{\sin(\pi y)}=\lim_{y\to 0}\frac{\pi y}{\sin(\pi y)}\cdot\frac{2-y}{\pi}=\color{red}{\frac{2}{\pi}}.$$