The limit of left{frac{1}{x} sqrt{1+x}-sqrt{1 | vedclass

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

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is equal to $\frac{1}{2}$

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(B) Let $L = \lim _{x \rightarrow 0} \left\{\frac{\sqrt{1+x}}{x} - \sqrt{1+\frac{1}{x^{2}}}\right\}$ $L = \lim _{x \rightarrow 0} \left\{\frac{\sqrt{1+x} - \sqrt{x^{2}+1}}{x}\right\}$ Since the limit is in the $\frac{0}{0}$ form,we apply $L$-Hospital's Rule: $L = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(\sqrt{1+x} - \sqrt{x^{2}+1})}{\frac{d}{dx}(x)}$ $L = \lim _{x \rightarrow 0} \frac{\frac{1}{2\sqrt{1+x}} - \frac{2x}{2\sqrt{x^{2}+1}}}{1}$ $L = \lim _{x \rightarrow 0} \left( \frac{1}{2\sqrt{1+x}} - \frac{x}{\sqrt{x^{2}+1}} \right)$ Substituting $x = 0$: $L = \frac{1}{2\sqrt{1+0}} - \frac{0}{\sqrt{0^{2}+1}} = \frac{1}{2} - 0 = \frac{1}{2}$

The limit of $\left\{\frac{1}{x} \sqrt{1+x}-\sqrt{1+\frac{1}{x^{2}}}\right\}$ as $x \rightarrow 0$ is:

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