Question

$\underset{x\to 0}{\lim }\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}$

Answer 1

$\underset{x\to 0}{\lim }\left[\frac{\sqrt{1-x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}\right]$

It is of the form $\frac{0}{0}$.
Rationalising the numerator and the denominator:

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{\left(\sqrt{1+x^2}-\sqrt{1+x}\right)}{1}\times \frac{\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}{\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}\times \frac{1}{\left(\sqrt{1+x^3}-\sqrt{1+x}\right)}\times \frac{\left(\sqrt{1+x^3}-\sqrt{1+x}\right)}{\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\left[\left(1+x^2\right)-\left(1+x\right)\right]}{\left[\left(1+x^3\right)-\left(1+x\right)\right]}\times \frac{\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}{\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}\right] \\ =\underset{x\to 0}{\lim }\left[\left(\frac{x^2-x}{x^3-x}\right)\times \frac{\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}{\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{x\left(x-1\right)}{x\left(x^2-1\right)}\frac{\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}{\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\left(x-1\right)\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}\right] \\ =\frac{\left(\sqrt{1+0}+\sqrt{1+0}\right)}{\left(0+1\right)\left(\sqrt{1+0}+\sqrt{1+0}\right)} \\ =\frac{2}{2} \\ =1 \end{gathered}$$