Limx→0sin5x.tan3xlog(1+x).(4x−1)=kloge4 then k is

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$\underset{x \to 0}{L i m} \frac{\sin 5 x . \tan 3 x}{\log (1 + x) . (4^{x} - 1)} = k {\log^{e}}_{4}$ then $k$ is

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$\frac{5}{4}$

$\frac{8}{4}$

$\frac{10}{4}$

answer is A.

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Detailed Solution

$\underset{x \to 0}{L i m} \frac{\sin 5 x . \tan 3 x}{\log (1 + x) . (4^{x} - 1)} = k \log_{4} e$

$\underset{x \to 0}{L i m} \frac{\frac{\sin 5 x}{x} . \frac{\tan 3 x}{x}}{\frac{\log (1 + x)}{x} . \frac{4^{x} - 1}{x}} = k \log_{4} e$

$\Rightarrow 15 \log_{4} e = k \log_{4} e$

$\Rightarrow k = 15$

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