How do you use the Squeeze Theorem to show that $sqrt (x) * e^(sin(pi/x))=0$ as x approaches zero?

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Answer 1 · 2015-10-11T17:47:56

$-1 <= sin(pi/x) <= 1$ for all $x != 0$ .

So, $e^-1 <= e^sin(pi/x) <= e^1$ for all $x != 0$ .
( $e^u$ is an increasing function)

For $x > 0$ , $sqrtx$ is defined and positive, so

$sqrtx/e <= sqrtxe^sin(pi/x) <= e sqrtx$

$lim_(xrarr0^+)sqrtx/e =0$ $" "$ and $" "$ $lim_(xrarr0^+)esqrtx =0$ .

Therefore, by the Squeeze Theorem,

$lim_(xrarr0^+)sqrtxe^sin(pi/x) =0$

How do you use the Squeeze Theorem to show that sqrt (x) * e^(sin(pi/x))=0sqrt (x) * e^(sin(pi/x))=0 as x approaches zero?