<s>[INST] 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...
 [/INST] Just in the same spirit as Paul, remember that, for small values of $y$, $$(1+y)^n=1+n y+\frac{1}{2} n(n-1) y^2+\frac{1}{6}n (n-1) (n-2)  y^3+O\left(y^4\right)$$ So, replace $n$ by $\frac{1}{3}$ and $y$ by $2x$ and obtain $$(1+2x)^{\frac{1}{3}}=1+\frac{2 x}{3}-\frac{4 x^2}{9}+\frac{40 x^3}{81}+O\left(x^4\right)$$ $$\frac{(1+2x)^{\frac{1}{3}}-1}{x}=\frac{2}{3}-\frac{4 x}{9}+\frac{40 x^2}{81}+O\left(x^3\right)$$ which shows the limit as well as the manner it is approached 
</s>