prove the limit?

Question

lim sqrt x * e^sin(pi/x) = 0, when x approach to 0+

Answer 1

For every x > 0, sin (pi/x) is in [-1, 1]. Since the exponencial function is continuous, it's bounded on [-1, -1], varies from 1/e to e. Thefore e^sin(pi/x) is bounded.

Since sqrt(x) -> 0 as x -> 0, it follows limit x -> 0+ sqrt(x) e^(sin(pi/x)) = 0

Answer 2

= ( lim sqrt(x) ) ( lim e^sin(pi/x) ) as x -> 0+
The function y = sqrt (x) is a parabola with vertex at (0,0) opening to the right with range [0, +infinitty). As x -> 0+, y -> 0.
The function y = sin (pi/x) does not converge to a single value as x -> 0+. Rather the value of y is bounded by [-1, 1] as x -> 0+. Therefore e^sin(pi/x) is bounded by [e^-1, e^1].
( lim sqrt(x) ) ( lim e^sin(pi/x) ) as x -> 0+
= (0) * [e^-1, e^1] = 0

Answer 3

your answer is 0