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...

I assume you're trying to avoid using calculus at all, as though you're trying to differentiate a function? Then let's try something completely different.
Let $u={\left(1+2x\right)}^{1/3}$. So $x=\frac{1}{2}\left(u^3-1\right)$, and $u \rightarrow 1$ as $x \rightarrow 0$. Then we have:
$$\lim_{x \rightarrow 0}{\frac{\left(1+2x\right)^{1/3}-1}{x}}=\lim_{u \rightarrow 1}{\frac{u-1}{\left(\frac{u^3-1}{2}\right)}}=2\lim_{u \rightarrow 1}{\frac{u-1}{u^3-1}}$$
And, niftily enough, you might know that $u^3-1=\left(u-1\right)\left(u^2+u+1\right)$ and hopefully you can see how useful that is.
