Evaluating $\lim\limits_{x\to1}\frac{\sqrt[3]{x^3+1}}{\sqrt[3]{x+1}}$? This is the limit:
$$\lim_{x\to1}\frac{\sqrt[\large 3]{x^3+1}}{\sqrt[\large 3]{x+1}}$$
I made the calculation and the result gave $\sqrt[\large 3]{0}$, I wonder if this is the correct result, but if not, what would be the account to the correct result. 
Thanks!

$$\lim_{x\to-1}\frac{\sqrt[\large 3]{x^3+1}}{\sqrt[\large 3]{x+1}}=\lim_{x\to-1}{\sqrt[\large 3]{\frac{x^3+1}{x+1}}}=\lim_{x\to-1}{\sqrt[\large 3]{\frac{(x+1)(x^2-x+1)}{x+1}}}=$$
$$=\lim_{x\to-1}{\sqrt[\large 3]{(x^2-x+1)}}=\sqrt[\large 3]3$$
