$Find $\lim_{x\rightarrow \infty} \frac{1}{x^2}\sin x^2\tan x$$

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$Find $\lim_{x\rightarrow \infty} \frac{1}{x^2}\sin x^2\tan x$$

How should I compute the limit
$$\lim_{x\rightarrow \infty} \frac{1}{x^2}\sin x^2\tan x?$$

Precalculus Limits Trigonometry

Daniel90

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Daniel90

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My teacher said it should be zero, but I don't know why.

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Daniel90

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My teacher said it should be zero, but I don't know why.

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First note that $ -1 \leq \sin x^2 \leq 1$. Thus
\[(1) 0=-\lim_{x \rightarrow \infty}\frac{1}{x^2} \tan x \leq \lim_{x \rightarrow \infty}\frac{1}{x^2} \sin x^2 \tan x \leq \lim_{x \rightarrow \infty}\frac{1}{x^2} \tan x=0,\]
and by the squeez theorem we have
\[\lim_{x \rightarrow \infty}\frac{1}{x^2} \sin x^2 \tan x=0.\]
Here we have used the following limit
\[\lim_{x \rightarrow \infty}\frac{1}{x^2} \tan x=\lim_{x \rightarrow \infty}\frac{\frac{1}{1+x^2}}{2x} =0,\]
which follows from the L'hopital's rule.

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$Find $\lim_{x\rightarrow \infty} \frac{1}{x^2}\sin x^2\tan x$$

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