Where is the mistake in this solution of $\lim_{x \to 1}{\frac{1-x^2}{\sin (\pi x)}}$?

Question

I'm trying to solve this limit:

$$\lim_{x \to 1}{\frac{1-x^2}{\sin (\pi x)}}$$

The answer ought to be $\frac{2}{\pi}$, but I end up with $0$:

$\lim\limits_{x \to 1}{\frac{1-x^2}{\sin (\pi x)}} = $ $\lim\limits_{y \to 0}{\frac{1-(y+1)^2}{\sin (\pi (y+1))}} = $ $\lim\limits_{y \to 0}{\frac{\pi(y+1)}{\sin (\pi (y+1))} \frac{1-(y+1)^2}{\pi(y+1)}} = $ $\lim\limits_{y \to 0}{\frac{1-(y+1)^2}{\pi(y+1)}} = 0$

Where and why is my solution incorrect?

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Note: I'm aware of this post, however I believe mine is different because I'm asking where and why my solution went wrong, not why my answer was wrong.

Answer 1

Your third equality attempts to make use of the rule $\lim\limits_{x\to0}\frac{x}{\sin x} = 1$, but note that yours has $y\to 0$ yet the argument is not $y$, it is $\pi(y+1)$, which does not go to zero. That's where your work goes wrong.

Answer 2

$$\lim _{ y\to 0 }{ \frac { \pi (y+1) }{ \sin (\pi (y+1)) } } =\frac { \pi }{ 0 } \neq 1\\ $$