Evaluate
$\displaystyle\lim_{x\to 0} \frac{\sqrt[3]{1+cx}-1}{x}$ where $c$ is a
constant.
Solution.
Introduce a new variable $t$ by
t=\sqrt[3]{1+cx}
We also need to express $x$ in terms of $t$:
\begin{eqnarray}
t^3=1+cx \\
\Longrightarrow x=\frac{t^3-1}{c}
\end{eqnarray}
As $x\to 0$, then $t\to 1$. We can therefore write
\begin{eqnarray}
\lim_{x\to 0}\frac{\sqrt[3]{1+cx}-1}{x} &=& \lim_{t\to
1}\frac{t-1}{(t^3-1)/c} \\
&=& \lim_{t\to 1}\frac{c(t-1)}{t^3-1} \\
&=& \lim_{t\to 1}\frac{c(t-1)}{(t-1)(t^2+t+1)} \\
&=& \lim_{t\to 1}\frac{c}{t^2+t+1} = \frac{c}3.
\end{eqnarray}