122. Find $\lim _{x \rightarrow 0} \frac{\sqrt{2+x}-\sqrt{2-x}}{5 x}$.
Solution. Here the limits of the numerator and denominator as $x \rightarrow 0$ are both zero. By multiplying the numerator and denominator by the expression conjugate to the numerator, we get

$$
\begin{gathered}
\frac{\sqrt{2+x}-\sqrt{2-x}}{5 x}=\frac{(\sqrt{2+x}-\sqrt{2-x})(\sqrt{2+x}+\sqrt{2-x})}{5 x(\sqrt{2+x}+\sqrt{2-x})}= \\
=\frac{2+x-2+x}{5 x(\sqrt{2+x}+\sqrt{2-x})}=\frac{2 x}{5 x(\sqrt{2+x}+\sqrt{2-x})}=\frac{2}{5(\sqrt{2+x}+\sqrt{2-x})}
\end{gathered}
$$

Therefore,

$$
\lim _{x \rightarrow 0} \frac{\sqrt{2+x}-\sqrt{2-x}}{5 x}=\lim _{x \rightarrow 0} \frac{2}{5(\sqrt{2+x}+\sqrt{2-x})}=
$$

$$
=\frac{\lim _{x \rightarrow 0} 2}{5 \lim _{x \rightarrow 0}(\sqrt{2+x}+\sqrt{2-x})}=\frac{2}{5 \cdot 2 \sqrt{2}}=\frac{\sqrt{2}}{10} .
$$