How do you find the limit of $x^2 sin(1/x)$ as x approaches 0?

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Answer 1 · 2016-08-30T17:19:21

Use the squeeze theorem.

We know from trigonometry that

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

Important: for $lim_(xrarr0)$ we don't care what happens when $x = 0$ .

Since $x<2 > 0$ for all $x != 0$ , we can multiply through by $x^2$ to get

$-x^2 = x^2 sin(1/x) <= x^2$ .

Clearly $lim_(xrarr0) (-x^2) = 0$ and $lim_(xrarr0) x^2 = 0$ , so, by the squeeze theorem,

$lim_(xrarr0) x^2 sin(1/x) = 0$ .

How do you find the limit of x^2 sin(1/x) x^2 sin(1/x) as x approaches 0?