lim _{x rightarrow 1} frac{log x}{1-x} = | vedclass

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(B) આપેલ લક્ષ $\lim _{x ightarrow 1} \frac{\log x}{1-x}$ છે.$x = 1$ મૂકતા,આપણને $\frac{\log 1}{1-1} = \frac{0}{0}$ સ્વરૂપ મળે છે.$L'	ext{Hospital's

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) આપેલ લક્ષ $\lim _{x ightarrow 1} \frac{\log x}{1-x}$ છે.$x = 1$ મૂકતા,આપણને $\frac{\log 1}{1-1} = \frac{0}{0}$ સ્વરૂપ મળે છે.$L'	ext{Hospital's Rule}$ નો ઉપયોગ કરતા,આપણે અંશ અને છેદનું $x$ ની સાપેક્ષમાં વિકલન કરીએ છીએ:$\lim _{x ightarrow 1} \frac{\frac{d}{dx}(\log x)}{\frac{d}{dx}(1-x)} = \lim _{x ightarrow 1} \frac{\frac{1}{x}}{-1}$.$x ightarrow 1$ લેતા:$\frac{\frac{1}{1}}{-1} = -1$.

$\lim _{x \rightarrow 1} \frac{\log x}{1-x} = $

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