# How do you use the Squeeze Theorem to find lim(x-1)sin(pi/x-1)  as x approaches one?

Oct 11, 2015

${\lim}_{x \rightarrow 1} \left(x - 1\right) \sin \left(\frac{\pi}{x} - 1\right) = 0$

#### Explanation:

Firstly, we don't need the squeeze theorem because the function is continuous at values close to $x = 1$, so the limit is $f \left(1\right)$, that being said, to use the squeeze theorem we must remember that

$- 1 \le \sin \left(\theta\right) \le 1$

If we have $\theta = \frac{\pi}{x} - 1$, we have

$- 1 \le \sin \left(\frac{\pi}{x} - 1\right) \le 1$

Multiplying both sides by $x - 1$ we have

$1 - x \le \left(x - 1\right) \sin \left(\frac{\pi}{x} - 1\right) \le x - 1$ for $x > 0$

Since

${\lim}_{x \rightarrow 1} 1 - x = {\lim}_{x \rightarrow 1} x - 1 = 0$

The squeeze theorem tells us that

${\lim}_{x \rightarrow 1} \left(x - 1\right) \sin \left(\frac{\pi}{x} - 1\right) = 0$

Oct 11, 2015

${\lim}_{x \rightarrow 1} \left(x - 1\right) \sin \left(\frac{\pi}{x - 1}\right) = 0$

#### Explanation:

$- 1 \le \sin \left(\frac{\pi}{x - 1}\right) \le 1$ for all $x \ne 1$

Limit from the right is 0

For $x > 1$, we have $x - 1 > 0$ so we can multiply the inequality without changing the inequalities:

$- \left(x - 1\right) \le \left(x - 1\right) \sin \left(\frac{\pi}{x - 1}\right) \le x - 1$ for all $x \ne 1$

${\lim}_{x \rightarrow {1}^{+}} - \left(x - 1\right) = 0$ and ${\lim}_{x \rightarrow {1}^{+}} \left(x - 1\right) = 0$

So ${\lim}_{x \rightarrow {1}^{+}} \left(x - 1\right) \sin \left(\frac{\pi}{x - 1}\right) = 0$

Limit from the left is 0

For $x < 1$, we have $x - 1 < 0$ so when we multiply the inequality we must change the inequalities:

$- \left(x - 1\right) \ge \left(x - 1\right) \sin \left(\frac{\pi}{x - 1}\right) \ge x - 1$ for all $x \ne 1$

${\lim}_{x \rightarrow {1}^{-}} - \left(x - 1\right) = 0$ and ${\lim}_{x \rightarrow {1}^{-}} \left(x - 1\right) = 0$

So ${\lim}_{x \rightarrow {1}^{-}} \left(x - 1\right) \sin \left(\frac{\pi}{x - 1}\right) = 0$

Therefore,

${\lim}_{x \rightarrow 1} \left(x - 1\right) \sin \left(\frac{\pi}{x - 1}\right) = 0$