155. Find $\lim _{x \rightarrow 0}\left(\frac{3+x}{3}\right)^{1 / x}$.
Solution. Write the base of the power as $\frac{3+x}{3}=1+\frac{x}{3}$, and the exponent as $\frac{1}{x}=\frac{1}{x} \cdot \frac{3}{3}=\frac{3}{x} \cdot \frac{1}{3}$. Therefore,

$$
\lim _{x \rightarrow 0}\left(\frac{3+x}{3}\right)^{1 / x}=\lim _{x \rightarrow 0}\left(1+\frac{x}{3}\right)^{3 / x \cdot 1 / 3}=\left(\lim _{x \rightarrow 0}\left(1+\frac{x}{3}\right)^{3 / x}\right)^{1 / 3}=e^{1 / 3}=\sqrt[3]{e}
$$