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.

Use definition of derivation:
\begin{align}
\frac{\sqrt{1+x^2} - \sqrt {1+x}}{\sqrt {1+x^3} - \sqrt {{1+x}}}
&=
\frac{\frac{\sqrt{1+x^2}-1}{x} - \frac{\sqrt {1+x}-1}{x}}{\frac{\sqrt {1+x^3}-1}{x} - \frac{\sqrt {{1+x}}-1}{x}}\\
&=
\frac{(\sqrt{1+x^2})'|_{x=0} - (\sqrt {1+x})'|_{x=0} }{(\sqrt {1+x^3})'|_{x=0}  - (\sqrt {{1+x}})'|_{x=0} }\\
&=
\dfrac{0-\frac12}{0-\frac12}\\
&=1
\end{align}
