<s>[INST] How do you evaluate limit of $\frac{\sqrt{1+x^2} - \sqrt {1+x}}{\sqrt {1+x^3} - \sqrt {{1+x}}}$ when $x$ tends to $0$? I tried rationalization method and got as $\frac{x^2-x}{\sqrt {1+x^3} - \sqrt {1+x} ({\sqrt{1+x^2} + \sqrt {1+x}})}$. But i feel the denominator having power of 3 I may be doing it wrong.The answer should be 1. Please help.
 [/INST] Use the fact that when $u\to 0$, $(1+u)^n \approx 1+nu$.
Using that fact, 
$$\begin{align} L &= \lim_{x\to 0} \frac{\sqrt{1+x^2} - \sqrt {1+x}}{\sqrt {1+x^3} - \sqrt {{1+x}}} \\ &= \lim_{x\to 0} \dfrac{1+\dfrac 12 x^2 - (1+\dfrac 12 x)}{1+\dfrac 12 x^3 - (1+\dfrac 12 x)}\\ &= \lim_{x\to 0} \dfrac{x^2-x}{x^3-x} \\ &= \lim_{x\to 0}\dfrac{1}{1+x} \\ &= \color{red}{1}\end{align}$$
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