# How do you use the Squeeze Theorem to find lim xcos(50pi/x) as x approaches zero?

Oct 15, 2015

See the explanation.

#### Explanation:

From trigonometry $- 1 \le \cos \theta \le 1$ for all real numbers $\theta$.

So,

$- 1 \le \cos \left(\frac{50 \pi}{x}\right) \le 1$ for all $x \ne 0$

From the right
For $x > 0$, we can multiply without changing the directions of the inequalities, so we get:

$- x \le \cos \left(\frac{50 \pi}{x}\right) \le x$ for $x > 0$.

Observe that , ${\lim}_{x \rightarrow {0}^{+}} \left(- x\right) = {\lim}_{x \rightarrow {0}^{+}} \left(x\right) = 0$,
so, ${\lim}_{x \rightarrow {0}^{+}} \cos \left(\frac{50 \pi}{x}\right) = 0$

From the left
For $x < 0$, when we multiply we must change the directions of the inequalities, so we get:

$- x \ge \cos \left(\frac{50 \pi}{x}\right) \ge x$ for $x < 0$.

Observe that , ${\lim}_{x \rightarrow {0}^{-}} \left(- x\right) = {\lim}_{x \rightarrow {0}^{-}} \left(x\right) = 0$,
so, ${\lim}_{x \rightarrow {0}^{-}} \cos \left(\frac{50 \pi}{x}\right) = 0$

Two-sided Limit

Because both the left and rights limits are $0$, we conclude that:

${\lim}_{x \rightarrow 0} \cos \left(\frac{50 \pi}{x}\right) = 0$