<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] $\sqrt{1+x^n}=1+x^n/2+O(x^{2n})$ as $x\to0$.
In particular $\sqrt{1+x}=1+x/2+O(x^2)$, $\sqrt{1+x^2}=1+O(x^2)$
and $\sqrt{1+x^3}=1+O(x^2)$. Then your fraction is
$$\frac{1+O(x^2)-(1+x/2+O(x^2))}{1+O(x^2)-(1+x/2+O(x^2))}
=\frac{-x/2+O(x^2)}{-x/2+O(x^2)}$$
etc.
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