lim x 0 ( ( 1 - cos 2 ( 3 x ) cos 3 ( 4 x ) ) ( sin 3 ( 4 x ) ( log e ( 2 x + 1 ) ) 5 ) ) is equal to [2023]

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Solution

Q.
$\lim_{x \to 0} ((\frac{1 - \cos^{2} (3 x)}{\cos^{3} (4 x)}) (\frac{\sin^{3} (4 x)}{(\log_{e} {(2 x + 1))}^{5}}))$ is equal to [2023]

1 24

2 15

3 9

4 18

Ans.
(4)

$\lim_{x \to 0} [(\frac{(1 - \cos^{2} 3 x)}{\cos^{3} (4 x)}) (\frac{\sin^{3} (4 x)}{(\log_{e} {(2 x + 1))}^{5}})]$

$= \lim_{x \to 0} [\frac{\sin^{2} (3 x)}{\cos^{3} (4 x)} \times \frac{\sin^{3} (4 x)}{[\log_{e} (2 x + 1)]^{5}}]$

$= \lim_{x \to 0} [\frac{\frac{\sin^{2} (3 x)}{{(3 x)}^{2}} \times \frac{\sin^{3} (4 x)}{{(4 x)}^{3}} \times {(3 x)}^{2} \times {(4 x)}^{3}}{\cos^{3} (4 x) \cdot {[\frac{\log_{e} (2 x + 1)}{2 x}]}^{5} \times {(2 x)}^{5}}]$

$= \frac{9 \times 64}{32} = 18$

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