How to evaluate the limit $\lim_{x\to 0} ((1+2x)^{1/3 }-1)/ x$ without using the l'Hospital's rule? $$\lim_{x\to 0} \frac{(1+2x)^{\frac{1}{3}}-1}{x}.$$
Please do not use the l'hospital's rule as I am trying to solve this limit without using that rule... to no avail...

Hint: Put $2x+1 = y^{3}$. And note that as $x \to 0 \implies y \to 1$. Then the limit becomes
\begin{align*}
\lim_{x \to 0} \frac{(2x+1)^{1/3}-1}{x} &= \lim_{y \to 1} \frac{y-1}{\frac{y^{3}-1}{2}}\\
&= \lim_{y \to 1} \frac{2}{y^{3}-1} \times (y-1)\\
&= \lim_{y \to 1} \frac{2}{y^{2}+y+1} = \frac{2}{3}
\end{align*}
