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$\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{(1-\sqrt{x})}{(1-x)}}$ is equal to
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The correct answer is:
$\sqrt{\frac{2}{3}}$
We have, $\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1-\sqrt{x}}{1-x}}$
$=\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1-\sqrt{x}}{(1+\sqrt x)(1- \sqrt x)}}$
$=\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1}{1+\sqrt{x}}}$
$=\left(\frac{1+1}{2+1}\right)^{\frac{1}{1+1}}=\left(\frac{2}{3}\right)^{\frac{1}{2}}=\sqrt{\frac{2}{3}}$
$=\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1-\sqrt{x}}{(1+\sqrt x)(1- \sqrt x)}}$
$=\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1}{1+\sqrt{x}}}$
$=\left(\frac{1+1}{2+1}\right)^{\frac{1}{1+1}}=\left(\frac{2}{3}\right)^{\frac{1}{2}}=\sqrt{\frac{2}{3}}$
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