lim _{x rightarrow 0} left[ frac{1}{x} - frac | vedclass

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(D) Given limit: $L = \lim _{x \rightarrow 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} \right] = \lim _{x \rightarrow 0} \left[ \frac{e^x - 1 - x}{x(e^x

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$\frac{1}{2}$

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(D) Given limit: $L = \lim _{x ightarrow 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} ight] = \lim _{x ightarrow 0} \left[ \frac{e^x - 1 - x}{x(e^x - 1)} ight]$ This is in the $\frac{0}{0}$ indeterminate form. Applying $L'	ext{Hopital's rule}$: $= \lim _{x ightarrow 0} \frac{\frac{d}{dx}(e^x - 1 - x)}{\frac{d}{dx}(xe^x - x)} = \lim _{x ightarrow 0} \frac{e^x - 1}{e^x + xe^x - 1}$ This is still in the $\frac{0}{0}$ form. Applying $L'	ext{Hopital's rule}$ again: $= \lim _{x ightarrow 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(e^x + xe^x - 1)} = \lim _{x ightarrow 0} \frac{e^x}{e^x + e^x + xe^x} = \lim _{x ightarrow 0} \frac{e^x}{2e^x + xe^x}$ Substituting $x = 0$: $= \frac{e^0}{2e^0 + 0 \cdot e^0} = \frac{1}{2 + 0} = \frac{1}{2}$

$\lim _{x \rightarrow 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} \right] = $

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