JEE MAIN - Mathematics (2023 - 8th April Morning Shift - No. 4)
$$\lim_\limits{x \rightarrow 0}\left(\left(\frac{\left(1-\cos ^{2}(3 x)\right.}{\cos ^{3}(4 x)}\right)\left(\frac{\sin ^{3}(4 x)}{\left(\log _{e}(2 x+1)\right)^{5}}\right)\right)$$ is equal to _____________.
15
18
9
24
Explanation
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left[\left(\frac{1-\cos ^2(3 x)}{\cos ^3(4 x)}\right)\left(\frac{\sin ^3(4 x)}{\left(\log _e(2 x+1)\right)^5}\right)\right] \\\\
& =\lim _{x \rightarrow 0}\left[\frac{1-\cos ^2(3 x)}{9 x^2} \times \frac{9 x^2}{\cos ^3(4 x)}\right] \times \frac{\frac{\sin ^3 4 x}{(4 x)^3} \times 64 x^3}{\left[\frac{\log _e(2 x+1)}{2 x}\right]^5 \times(2 x)^5} \\\\
& =\left[\frac{1 \times 9 \times 1}{(1)}\right] \times\left[\frac{1 \times 64}{1 \times 32}\right] \\\\
& =9 \times 2=18
\end{aligned}
$$
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