જો log (1+x)=x-frac{x^2}{2}+frac{x^3}{3}-frac | vedclass

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(C) આપેલ લક્ષનું પદ: $\lim _{x \rightarrow 0} \left[ \frac{\log (1+x)^{1+x}}{x^2} - \frac{1}{x} \right] = k$ ગુણધર્મ $\log(a^b) = b \log a$ નો ઉપયોગ ક

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) આપેલ લક્ષનું પદ: $\lim _{x ightarrow 0} \left[ \frac{\log (1+x)^{1+x}}{x^2} - \frac{1}{x} ight] = k$ ગુણધર્મ $\log(a^b) = b \log a$ નો ઉપયોગ કરતા: $\lim _{x ightarrow 0} \left[ \frac{(1+x) \log (1+x) - x}{x^2} ight] = k$ આ $\frac{0}{0}$ સ્વરૂપ હોવાથી,આપણે $L$' Hospital નો નિયમ વાપરીએ: $\lim _{x ightarrow 0} \frac{\frac{d}{dx} [(1+x) \log (1+x) - x]}{\frac{d}{dx} [x^2]} = k$ $\lim _{x ightarrow 0} \frac{[1 \cdot \log (1+x) + (1+x) \cdot \frac{1}{1+x}] - 1}{2x} = k$ $\lim _{x ightarrow 0} \frac{\log (1+x) + 1 - 1}{2x} = k$ $\lim _{x ightarrow 0} \frac{\log (1+x)}{2x} = k$ $\frac{1}{2} \lim _{x ightarrow 0} \frac{\log (1+x)}{x} = k$ કારણ કે $\lim _{x ightarrow 0} \frac{\log (1+x)}{x} = 1$,તેથી $k = \frac{1}{2} 	imes 1 = \frac{1}{2}$. તેથી,$12k = 12 	imes \frac{1}{2} = 6$.

જો $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \infty$ અને $\lim _{x \rightarrow 0} \frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}=k$ હોય,તો $12 k=$

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