Find the value

Solution:

To evaluate:

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{2}}-\sqrt{1+x}}{\sqrt{1+x^{3}}-\sqrt{1+x}}$

Formula used: L'Hospital's rule

Let $f(x)$ and $g(x)$ be two functions which are differentiable on an open interval I except at a point a where

$\lim _{x \rightarrow a} \mathrm{f}(\mathrm{x})=\lim _{x \rightarrow a} \mathrm{~g}(\mathrm{x})=0$ or $\pm \infty$

then

$\lim _{x \rightarrow a} \frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}=\lim _{x \rightarrow a} \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{g}^{\prime}(\mathrm{x})}$

As $x \rightarrow 0$, we have

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{2}}-\sqrt{1+x}}{\sqrt{1+x^{3}}-\sqrt{1+x}}=\frac{0}{0}$

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{2}}-\sqrt{1+x}}{\sqrt{1+x^{3}}-\sqrt{1+x}}=\lim _{x \rightarrow 0} \frac{\frac{d}{d x}\left(\sqrt{1+x^{2}}-\sqrt{1+x}\right)}{\frac{d}{d x}\left(\sqrt{1+x^{3}}-\sqrt{1+x}\right)}$

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{2}}-\sqrt{1+x}}{\sqrt{1+x^{3}}-\sqrt{1+x}}=\lim _{x \rightarrow 0} \frac{\frac{2 x}{2 \sqrt{1+x^{2}}}-\frac{1}{2 \sqrt{1+x}}}{\frac{3 x^{2}}{2 \sqrt{1+x^{3}}}-\frac{1}{2 \sqrt{1-x}}}$

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{2}}-\sqrt{1+x}}{\sqrt{1+x^{3}}-\sqrt{1+x}}=\lim _{x \rightarrow 0} \frac{-\frac{1}{2}}{-\frac{1}{2}}$

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{2}}-\sqrt{1+x}}{\sqrt{1+x^{3}}-\sqrt{1+x}}=-1$

Thus, the value of $\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{2}}-\sqrt{1+x}}{\sqrt{1+x^{3}}-\sqrt{1+x}}$ is $-1$.