Question #962f6

Question

4607 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-15T18:37:19

Use the squeeze (pinch, sandwich) theorem (right limit version).

In order to stay in the real numbers, I assume that we want $lim_(xrarr0^+) sqrt(x) sin(1/x)$ .

Note the for all $x != 0$ , we have

$-1 <= sin(1/x) <= 1$

Since $sqrtx > 0$ , we can multiply through by $sqrtx$ without changing the inequalities.

$-sqrtx <= sqrtx sin(1/x) <= sqrtx$ .

Of, course, $lim_(xrarr0^+) -sqrtx = 0 = lim_(xrarr0^+) sqrtx$ .

By the squeeze theorem, $lim_(xrarr0^+) sqrtx sin(1/x) = 0$