Evaluate

Solution:

To evaluate:

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

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 $\mathrm{X} \rightarrow 0$, we have

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

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

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

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

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

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