Question

$\lim_{x \to 0^{+}} \sqrt{x} e^{\sin(\frac{1}{x})}$

Answer 1

0

Answer 2

The limit is evaluated by recognizing that one part of the product ($\sqrt{x}$) tends to zero, while the other part ($e^{\sin(1/x)}$) is bounded. Specifically, $\sin(1/x)$ oscillates between -1 and 1, so $e^{\sin(1/x)}$ oscillates between $e^{-1}$ and $e^1$. By the Squeeze Theorem, the product of a function tending to zero and a bounded function is zero.