(1) \( \lim _{x \rightarrow 0^{+}} \frac{\sin \sqrt{x}}{x}=\lim _{x \rightarrow 0^{-}} \)

To tackle the limit \( \lim _{x \rightarrow 0^{+}} \frac{\sin \sqrt{x}}{x} \), we start with the fact that as \( x \) approaches \( 0 \) from the positive side, \( \sqrt{x} \) also approaches \( 0 \). Using the small angle approximation, \( \sin y \approx y \) when \( y \) is near \( 0 \), we can set \( y = \sqrt{x} \). Thus, we rewrite the limit: \[ \lim _{x \rightarrow 0^{+}} \frac{\sin \sqrt{x}}{x} = \lim _{x \rightarrow 0^{+}} \frac{\sin \sqrt{x}}{\sqrt{x}} \cdot \frac{\sqrt{x}}{x} = \lim _{x \rightarrow 0^{+}} \frac{\sin \sqrt{x}}{\sqrt{x}} \cdot \frac{1}{\sqrt{x}} \] Here, the limit \( \lim _{x \rightarrow 0^{+}} \frac{\sin \sqrt{x}}{\sqrt{x}} = 1 \). Therefore, we have: \[ \lim _{x \rightarrow 0^{+}} \frac{\sin \sqrt{x}}{x} = \lim _{x \rightarrow 0^{+}} \frac{1}{\sqrt{x}} = \infty \] Thus, \( \lim _{x \rightarrow 0^{+}} \frac{\sin \sqrt{x}}{x} = \infty \) and similarly, since the left-hand limit is not applicable (as \( x \) cannot approach \( 0 \) from the negative side), we write that \( \lim _{x \rightarrow 0^{-}} \frac{\sin \sqrt{x}}{x} \) does not exist.