JEE Main 2023 — Mathematics Calculus

Given,

$$\underset{x\rightarrow 0}{\mathrm{lim}}((\frac{1-{\mathrm{cos}}^{2}(3x)}{{\mathrm{cos}}^{3}(4x)})(\frac{{\mathrm{sin}}^{3}(4x)}{{({\mathrm{log}}_{e}(2x+1))}^{5}}))$$x→0lim​((cos3(4x)1−cos2(3x)​)((loge​(2x+1))5sin3(4x)​))

Now we now that,

\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\mathrm{sin}x}{x}=1,\underset{x\rightarrow 0}{\mathrm{lim}}\frac{1-\mathrm{cos}x}{{x}^{2}}=\frac{1}{2}&\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\mathrm{log}(1+x)}{x}=1

Now using the above formula we get,

$$\underset{x\rightarrow 0}{\mathrm{lim}}((\frac{1-{\mathrm{cos}}^{2}3x}{{\mathrm{cos}}^{3}4x})(\frac{{\mathrm{sin}}^{3}4x}{{({\mathrm{log}}_{e}(2x+1))}^{5}}))$$x→0lim​((cos34x1−cos23x​)((loge​(2x+1))5sin34x​))

$$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{(1-\mathrm{cos}3x)(1+\mathrm{cos}3x)9{x}^{2}}{({\mathrm{cos}}^{3}4x)9{x}^{2}}\frac{{(\mathrm{sin}4x)}^{3}{(64x)}^{3}{(2x)}^{5}}{(64{x}^{3}){({\mathrm{log}}_{e}(2x+1))}^{5}{(2x)}^{5}}$$=x→0lim​(cos34x)9x2(1−cos3x)(1+cos3x)9x2​(64x3)(loge​(2x+1))5(2x)5(sin4x)3(64x)3(2x)5​

$$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{(\frac{1-\mathrm{cos}3x}{9{x}^{2}})(1+\mathrm{cos}3x)}{({\mathrm{cos}}^{3}4x)}\frac{{(\frac{\mathrm{sin}4x}{4x})}^{3}}{{(\frac{{\mathrm{log}}_{e}(2x+1)}{2x})}^{5}}\times \frac{9\times 64}{32}$$=x→0lim​(cos34x)(9x21−cos3x​)(1+cos3x)​(2xloge​(2x+1)​)5(4xsin4x​)3​×329×64​

$$=\frac{(\frac{1}{2})\times (2)}{(1)}\frac{{(1)}^{3}}{{(1)}^{5}}\times \frac{9\times 64}{32}$$=(1)(21​)×(2)​(1)5(1)3​×329×64​

$$=\frac{9\times 2}{1}=18$$=19×2​=18