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.
 A: Using the same rationalization method, when $x\not\in\{0,1\}$,
$$
\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}
=\frac1{x+1}\frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}
$$
A: 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}
A: Use the fact that when $u\to 0$, $(1+u)^n \approx 1+nu$.
Using that fact, 
$$\begin{align} L &= \lim_{x\to 0} \frac{\sqrt{1+x^2} - \sqrt {1+x}}{\sqrt {1+x^3} - \sqrt {{1+x}}} \\ &= \lim_{x\to 0} \dfrac{1+\dfrac 12 x^2 - (1+\dfrac 12 x)}{1+\dfrac 12 x^3 - (1+\dfrac 12 x)}\\ &= \lim_{x\to 0} \dfrac{x^2-x}{x^3-x} \\ &= \lim_{x\to 0}\dfrac{1}{1+x} \\ &= \color{red}{1}\end{align}$$
A: $\sqrt{1+x^n}=1+x^n/2+O(x^{2n})$ 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.
A: You should rationalize both numerator and denominator.
Terms will get cancelled and you will get the final answer
