Find the limit? How to calculate this limit ? 
$$\lim _{x\to 1}\frac{\sqrt{x^2-1}}{\sqrt[3]{x-1}}$$
An image for clarification.

We can rewrite the function:
$$\frac{\sqrt{x^{2}-1}}{\sqrt[3]{x-1}}=(x+1)^{\frac{1}{2}}(x-1)^{\frac{1}{2}}(x-1)^{-\frac{1}{3}}=(x+1)^{\frac{1}{2}}(x-1)^{\frac{1}{2}-\frac{1}{3}}$$
So that
$$\lim_{x\to 1}\frac{\sqrt{x^{2}-1}}{\sqrt[3]{x-1}}=\lim_{x\to 1}(x+1)^{\frac{1}{2}}(x-1)^{\frac{1}{6}}=0$$
