Question

Question #962f6

Answer 1

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

Explanation:

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`