# Where is the mistake in this solution of <munder> <mo movablelimits="true" form="prefix">lim

Where is the mistake in this solution of $\underset{x\to 1}{lim}\frac{1-{x}^{2}}{\mathrm{sin}\left(\pi x\right)}$?
The answer ought to be $\frac{2}{\pi }$, but I end up with 0:
$\underset{x\to 1}{lim}\frac{1-{x}^{2}}{\mathrm{sin}\left(\pi x\right)}=$ $\underset{y\to 0}{lim}\frac{1-\left(y+1{\right)}^{2}}{\mathrm{sin}\left(\pi \left(y+1\right)\right)}=$ $\underset{y\to 0}{lim}\frac{\pi \left(y+1\right)}{\mathrm{sin}\left(\pi \left(y+1\right)\right)}\frac{1-\left(y+1{\right)}^{2}}{\pi \left(y+1\right)}=$ $\underset{y\to 0}{lim}\frac{1-\left(y+1{\right)}^{2}}{\pi \left(y+1\right)}=0$
Where and why is my solution incorrect?
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livin4him777lf
Your third equality attempts to make use of the rule $\underset{x\to 0}{lim}\frac{x}{\mathrm{sin}x}=1$, but note that yours has $y\to 0$ yet the argument is not y, it is $\pi \left(y+1\right)$, which does not go to zero. That's where your work goes wrong.

Boilanubjaini8f
$\underset{y\to 0}{lim}\frac{\pi \left(y+1\right)}{\mathrm{sin}\left(\pi \left(y+1\right)\right)}=\frac{\pi }{0}\ne 1\phantom{\rule{0ex}{0ex}}$