Answer
$$ \lim _{h \rightarrow 0} \frac{\sqrt[3]{1+h}-1}{\sqrt[2]{1+h}-1}=\frac{2}{3}$$
Work Step by Step
Given $$ \lim _{h \rightarrow 0} \frac{\sqrt[3]{1+h}-1}{\sqrt[2]{1+h}-1}$$
let $$ f(h) = \frac{\sqrt[3]{1+h}-1}{\sqrt[2]{1+h}-1}$$
Since, we have
$$ f( 0)= \frac{1-1}{1-1}=\frac{0}{0}$$
Let $\sqrt[6]{1+h}=x \ $
$ \Rightarrow \sqrt[3]{1+h}=x^2,$
$ \Rightarrow \sqrt[2]{1+h}=x^3\\
h=x^6-1$
Also, when $ h\rightarrow 0, \ \ \Rightarrow x \rightarrow 1 $,
So, transform algebraically and cancel
\begin{aligned}
L&= \lim _{h \rightarrow 0} \frac{\sqrt[3]{1+h}-1}{\sqrt[2]{1+h}-1} \\
&=\lim _{x \rightarrow 1} \frac{x^{2}-1}{x^{3}-1} \\ &=\lim _{x \rightarrow 1} \frac{(x-1)(x+1)}{(x-1)\left(x^{2}+x+1\right)} \\ &=\lim _{x \rightarrow 1} \frac{x+1}{x^{2}+x+1} \\ &=\frac{1+1}{\left(1^{2}+1+1\right)} \\ &=\frac{2}{3} \
\end{aligned}
