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

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

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

Where and why is my solution incorrect?

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

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

Where and why is my solution incorrect?