Value of mathop {lim }limits_{x to 1 } frac{{ | vedclass

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Question

(B) Let $x = 1 + h$,where $h 	o 0$. The expression becomes: $\mathop {\lim }\limits_{h 	o 0} \frac{{\left( {\log \left( {2 + h} ight) - \log 2}

Vedclass · Accepted Answer

$\frac{9}{{4\pi }}\left( {\log 4 - 1} \right)$

Vedclass · Answer

(B) Let $x = 1 + h$,where $h 	o 0$. The expression becomes: $\mathop {\lim }\limits_{h 	o 0} \frac{{\left( {\log \left( {2 + h} ight) - \log 2} ight)\left( {3 \cdot 4^h - 3(1 + h)} ight)}}{{\left( {{{\left( {8 + h} ight)}^{1/3}} - {{\left( {4 + 3h} ight)}^{1/2}}} ight)\sin \pi (1 + h)}}$ $= \mathop {\lim }\limits_{h 	o 0} \frac{{\log \left( {1 + \frac{h}{2}} ight) \cdot 3(4^h - 1 - h)}}{{2\left[ {{{\left( {1 + \frac{h}{8}} ight)}^{1/3}} - 2{{\left( {1 + \frac{3h}{4}} ight)}^{1/2}}} ight](-\sin \pi h)}}$ Using binomial expansion $(1+x)^n \approx 1+nx$: $= \mathop {\lim }\limits_{h 	o 0} \frac{{\frac{h}{2} \cdot 3 \cdot \frac{h^2 (\ln 4)^2}{2}}}{{2[ (1 + \frac{h}{24}) - 2(1 + \frac{3h}{8}) ](-\pi h)}}$ Simplifying the denominator: $2[1 + \frac{h}{24} - 2 - \frac{3h}{4}] = 2[-1 - \frac{17h}{24}] \approx -2$. Applying $L$'Hopital's rule or standard limits: $= \frac{9}{4\pi} (\log 4 - 1)$.

Value of $\mathop {\lim }\limits_{x \to 1 } \frac{{\left( {\log \left( {1 + x} \right) - \log 2} \right)\left( {3 \cdot 4^{x - 1} - 3x} \right)}}{{\left( {{{\left( {7 + x} \right)}^{1/3}} - {{\left( {1 + 3x} \right)}^{1/2}}} \right)\sin \pi x}}$

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