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Q. $\displaystyle\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{2}}-\sqrt{1-x+x^{2}}}{3^{x}-1}$ is equal to

EAMCETEAMCET 2014

Solution:

$\displaystyle\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{2}}-\sqrt{1-x+x^{2}}}{3^{x}-1}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{2}}-\sqrt{1-x+x^{2}}}{3^{x}-1}\times \frac{\sqrt{1+x^{2}}+\sqrt{1-x+x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x+x^{2}}}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{\left(1+x^{2}\right)-\left(1-x+x^{2}\right)}{3^{x}-1\left(\sqrt{1-x^{2}}+\sqrt{1-x+x^{2}}\right)}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{x}{\left(3^{x}-1\right)} \times \frac{1}{\left(\sqrt{1-x^{2}}+\sqrt{1-x+x^{2}}\right)}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{1}{\frac{3^{x}-1}{x}} \times \frac{1}{\sqrt{1-x^{2}}+\sqrt{1-x+x^{2}}}$
$=\log _{e} \frac{1}{3} \times \frac{1}{\sqrt{1-0}+\sqrt{1-0+0}}\,\,\,\left[\because \log \frac{\left(a^{x}-1\right)}{x}=1\right]$
$=\frac{1}{2 \log _{e} 3}=\frac{1}{\log _{e} 9}$