JEE Main 2023 — Limits Question with Solution

Given,

$\underset{x\to 0}{\lim }\left(\left(\frac{1-{\cos }^2\left(3x\right)}{{\cos }^3\left(4x\right)}\right)\left(\frac{{\sin }^3\left(4x\right)}{{\left({\log }_e\left(2x+1\right)\right)}^5}\right)\right)$

Now we now that,

$\underset{x\to 0}{\lim }\frac{\sin x}{x}=1, \underset{x\to 0}{\lim }\frac{{\displaystyle 1-\cos x}}{{\displaystyle x^2}}=\frac{{\displaystyle 1}}{2} \& \underset{x\to 0}{\lim }\frac{{\displaystyle \log \left(1+x\right)}}{{\displaystyle x}}=1$

Now using the above formula we get,

$\underset{x\to 0}{\lim }\left(\left(\frac{1-{\cos }^{2}3x}{{\cos }^{3}4x}\right)\left(\frac{{\sin }^{3}4x}{{\left({\log }_e\left(2x+1\right)\right)}^5}\right)\right)$

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

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

$=\frac{\left({\displaystyle \frac{1}{2}}\right)\times \left(2\right)}{\left(1\right)}\frac{{\left({\displaystyle 1}\right)}^3}{{\left({\displaystyle 1}\right)}^5}\times \frac{9\times 64}{32}$

$=\frac{9\times 2}{1}=18$