Mohamed Mooney

2022-06-25

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?

livin4him777lf

Beginner2022-06-26Added 14 answers

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 $\pi (y+1)$, which does not go to zero. That's where your work goes wrong.

Boilanubjaini8f

Beginner2022-06-27Added 6 answers

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

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