For $f\left(x\right)$ to be continuous at $x=0,$ we must have
$f\left(0\right)=\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\sqrt[3]{1+2x}-\sqrt[4]{1+x}}{x}$
$=\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\left[1+\frac{2}{3}x+\frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2!}4x^2+.....\right]}{x}-\frac{\left[1+\frac{1}{4}x+\frac{\left(\frac{1}{4}\right)\left(-\frac{3}{4}\right)}{2!}{x}^2+.....\right]}{x}$
$=\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{x\left[\frac{5}{12}+\left(\frac{-4}{9}+\frac{3}{32}\right)x+\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s} \mathrm{c}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{a}\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{n}\mathrm{g} x^2 \mathrm{a}\mathrm{n}\mathrm{d} \mathrm{h}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{e}\mathrm{r} \mathrm{p}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}\mathrm{s}\right]}{x}$
$=\frac{5}{12}$

$\therefore f\left(0\right)=\frac{5}{12}$