JEE 2023 — Mathematics PYQ

$={\lim }_{x\to 0}\left[\frac{1-{\cos }^2(3x)}{9x^2}\times \frac{9x^2}{{\cos }^3(4x)}\right]\times \frac{\frac{{\sin }^3(4x)}{(4x{)}^3}\times 64x^3}{{\left[\frac{{\log }_e(2x+1)}{2x}\right]}^5\times (2x{)}^5}$
$=\left[\frac{1\times 9\times 1}{(1)}\right]\times \left[\frac{1\times 64}{1\times 32}\right]$
$=9\times 2=18$