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

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Answer 1 · 2015-10-15T21:51:08

See the explanation.

From trigonometry $-1 <= cos theta <=1$ for all real numbers $theta$ .

So,

$-1 <= cos ((50pi)/x) <= 1$ for all $x != 0$

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

$-x <= cos ((50pi)/x) <= x$ for $x > 0$ .

Observe that , $lim_(xrarr0^+) (-x) = lim_(xrarr0^+) (x) = 0$ ,
so, $lim_(xrarr0^+)cos ((50pi)/x) = 0$

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

$-x >= cos ((50pi)/x) >= x$ for $x < 0$ .

Observe that , $lim_(xrarr0^-) (-x) = lim_(xrarr0^-) (x) = 0$ ,
so, $lim_(xrarr0^-)cos ((50pi)/x) = 0$

Two-sided Limit

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

$lim_(xrarr0)cos ((50pi)/x) = 0$

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