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

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(D) આપેલ લક્ષ: $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 -

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(D) આપેલ લક્ષ: $L = \lim _{x ightarrow 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} ight] = \lim _{x ightarrow 0} \left[ \frac{e^x - 1 - x}{x(e^x - 1)} ight]$ આ $\frac{0}{0}$ અનિશ્ચિત સ્વરૂપમાં છે. $L'	ext{Hopital}$ નો નિયમ લાગુ પાડતા: $= \lim _{x ightarrow 0} \frac{\frac{d}{dx}(e^x - 1 - x)}{\frac{d}{dx}(xe^x - x)} = \lim _{x ightarrow 0} \frac{e^x - 1}{e^x + xe^x - 1}$ આ હજુ પણ $\frac{0}{0}$ સ્વરૂપમાં છે. ફરીથી $L'	ext{Hopital}$ નો નિયમ લાગુ પાડતા: $= \lim _{x ightarrow 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(e^x + xe^x - 1)} = \lim _{x ightarrow 0} \frac{e^x}{e^x + e^x + xe^x} = \lim _{x ightarrow 0} \frac{e^x}{2e^x + xe^x}$ $x = 0$ મૂકતા: $= \frac{e^0}{2e^0 + 0 \cdot e^0} = \frac{1}{2 + 0} = \frac{1}{2}$

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

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