Question #962f6

1 Answer
Nov 15, 2016

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#

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