I currently have $\lim_{h\to 0}\frac{\sqrt[3]{h+1}-1}{h}$
Now, I know that when you have square root instead of a cubic root it's easy. You just multiply by the conjugate and simplify afterwards. If it were a sqrt I know that the limit is 1/2. I know that the result here is 1/3 but I can't seem to get there. I always end up with 1/4 due to getting something like: $\frac{h+1-1}{h(\sqrt[3]{h+1}+1)(\sqrt[3]{h+1}+1)}$
After simplifying and evaluating the limit you end up with $\frac{1}{(\sqrt[3]{0+1}+1)(\sqrt[3]{0+1}+1)}$
which turns out to be $1/4$.
What am I doing wrong?