What is the limit of $\frac{{sin}^{4} (x)}{x^{0.5}}$ $sin^4(x)/x^0.5$ as x goes to infinity?

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Answer 1 · 2015-10-16T17:34:47

It is $0$ $0$

$0 \leq {sin}^{4} x \leq 1$ $0 <= sin^4x <= 1$ for all $x$ $x$

$x^{0.5} > 0$ $x^0.5 > 0$ for $x > 0$ $x > 0$ , so we can divide in the inequality without changing the directions of the inequalities.

$0 \leq \frac{{sin}^{4} x}{x^{0.5}} \leq \frac{1}{x^{0.5}}$ $0 <= sin^4x/x^0.5 <= 1/x^0.5$ for all $x$ $x$

We note that $lim_(xrarroo) 0 = lim_(xrarroo) 1/x^0.5 = 0$ ,

so, by the squeeze theorem, $lim_(xrarroo) sin^4x/x^0.5 = 0$

What is the limit of sin^4(x)/x^0.5sin4(x)x0.5sin^4(x)/x^0.5 as x goes to infinity?