$\underset{x\to a}{lim} \frac{(a+2x{)}^{1/3}-(3x{)}^{1/3}}{(3a+x{)}^{1/3}-(4x{)}^{1/3}},(a\ne 0)$ is equal to

Let $L=\underset{x\to a}{lim} \frac{(a+2x{)}^{1/3}-(3x{)}^{1/3}}{(3a+x{)}^{1/3}-(4x{)}^{1/3}}.$ Applying $L\text{'}$

Hospital's rule, we obtain 

$\begin{array}{l}L=\underset{x\to a}{lim} \frac{\frac{2}{3}(a+2x{)}^{-2/3}-(3x{)}^{-2/3}}{\frac{1}{3}(3a+x{)}^{-2/3}-\frac{4}{3}(4x{)}^{-2/3}}1\\ \Rightarrow L=\frac{-\frac{1}{3}(3a{)}^{-2/3}}{-(4a{)}^{-2/3}}\\ \Rightarrow L=\frac{1}{3}\frac{(4a{)}^{2/3}}{(3a{)}^{2/3}}=\frac{2^{4/3}}{3^{5/3}}=\frac{2}{3}{\left(\frac{2}{9}\right)}^{1/3}\end{array}$