The value of $f(0)$ so that the function $f(x) = \frac{\sqrt{1 + x} - \sqrt[3]{1 + x}}{x}$ becomes continuous is equal to

For continuity at $x=0$, we need $f(0)=\lim_{x\to 0} \frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}$. Using the expansions near $x=0$, $$\sqrt{1+x}=1+\frac{x}{2}+\cdots$$ and $$\sqrt[3]{1+x}=1+\frac{x}{3}+\cdots$$ So the numerator becomes $$\left(1+\frac{x}{2}\right)-\left(1+\frac{x}{3}\right)=\frac{x}{6}$$ Hence, $$\lim_{x\to 0} \frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}=\frac{1}{6}$$ Therefore the required value is $\frac{1}{6}$. Comparing with the given options, this matches option $\text{A}$.