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

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

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Answer 1 · 2015-10-11T17:47:56

$- 1 \leq sin (\frac{π}{x}) \leq 1$ $-1 <= sin(pi/x) <= 1$ for all $x \neq 0$ $x != 0$ .

So, $e^{- 1} \leq e^{sin (\frac{π}{x})} \leq e^{1}$ $e^-1 <= e^sin(pi/x) <= e^1$ for all $x \neq 0$ $x != 0$ .
( $e^{u}$ $e^u$ is an increasing function)

For $x > 0$ $x > 0$ , $\sqrt{x}$ $sqrtx$ is defined and positive, so

$\frac{\sqrt{x}}{e} \leq \sqrt{x} e^{sin (\frac{π}{x})} \leq e \sqrt{x}$ $sqrtx/e <= sqrtxe^sin(pi/x) <= e sqrtx$

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

Therefore, by the Squeeze Theorem,

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

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

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