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The limit of $\left\{\frac{1}{x} \sqrt{1+x}-\sqrt{1+\frac{1}{x^{2}}}\right\}$ as $x \rightarrow 0$
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is equal to $\frac{1}{2}$
$\lim _{x \rightarrow 0}\left\{\frac{\sqrt{1+x}}{x}-\sqrt{1+\frac{1}{x^{2}}}\right\}$
$=\lim _{x \rightarrow 0}\left\{\frac{\sqrt{1+x}-\sqrt{1+x^{2}}}{x}\right\}\left(\frac{0}{0}\right.$ form $)$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{2 \sqrt{1+x}}-\frac{x}{\sqrt{1+x^{2}}}}{1}$ (Use L-Hospital Rule)
$=\frac{1}{2 \sqrt{1+0}}-0=\frac{1}{2}$
$=\lim _{x \rightarrow 0}\left\{\frac{\sqrt{1+x}-\sqrt{1+x^{2}}}{x}\right\}\left(\frac{0}{0}\right.$ form $)$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{2 \sqrt{1+x}}-\frac{x}{\sqrt{1+x^{2}}}}{1}$ (Use L-Hospital Rule)
$=\frac{1}{2 \sqrt{1+0}}-0=\frac{1}{2}$
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