Question #5d46e

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Answer 1 · 2017-02-09T18:11:50

The limit is $0$ . Use the squeeze theorem.

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

For $x > 0$ and near $0$ , we have $sinx > 0$ so we can multiply the inequality by $sinx$ to get

$-sinx <= sinx sin(1/x) <= sinx$ for $0 < x <+ pi/2$

Observe that $lim_(xrarr0^+) -sinx = 0$ and lim_(xrarr0^+) -sinx =0#,

therefore, $lim_(xrarr0^+) sinxsin(1/x) = 0$

For $x < 0$ and near $0$ , we have sinx < 0#, so when we multiply the inequality we must revese the directions:

$-sinx >= sinxsin(1/x) >= sinx$

Observe that $lim_(xrarr0^-) -sinx = 0$ and lim_(xrarr0^-) -sinx =0#,

therefore, $lim_(xrarr0^-) sinxsin(1/x) = 0$ .

Since both one-sided limits are $0$ , the two-sided limit is also $0$ .