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