JEE Main 2022 — Mathematics Calculus

Consider $\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\mathrm{cos}(\mathrm{sin}x)-\mathrm{cos}x}{{x}^{4}}(\frac{0}{0}\text{form})$x→0lim​x4cos(sinx)−cosx​(00​form)

$$\underset{x\rightarrow 0}{\mathrm{lim}}\frac{2\mathrm{sin}(\frac{\mathrm{sin}x+x}{2})\cdot \mathrm{sin}(\frac{x-\mathrm{sin}x}{2})}{{x}^{4}}$$x→0lim​x42sin(2sinx+x​)⋅sin(2x−sinx​)​

$$=\underset{x\rightarrow 0}{\mathrm{lim}}2[\frac{\mathrm{sin}(\frac{\mathrm{sin}x+x}{2})}{(\frac{\mathrm{sin}x+x}{2})}][\frac{\mathrm{sin}(\frac{x-\mathrm{sin}x}{2})}{(\frac{x-\mathrm{sin}x}{2})}]\times (\frac{\mathrm{sin}x+x}{2})\times (\frac{x-\mathrm{sin}x}{2})\times \frac{1}{{x}^{4}}$$=x→0lim​2[(2sinx+x​)sin(2sinx+x​)​][(2x−sinx​)sin(2x−sinx​)​]×(2sinx+x​)×(2x−sinx​)×x41​

$$\underset{x\rightarrow 0}{\mathrm{lim}}2[\frac{\mathrm{sin}(\frac{\mathrm{sin}x+x}{2})}{(\frac{\mathrm{sin}x+x}{2})}][\frac{\mathrm{sin}(\frac{x-\mathrm{sin}x}{2})}{(\frac{x-\mathrm{sin}x}{2})}]\times (\frac{{x}^{2}-{\mathrm{sin}}^{2}x}{4{x}^{4}})$$x→0lim​2[(2sinx+x​)sin(2sinx+x​)​][(2x−sinx​)sin(2x−sinx​)​]×(4x4x2−sin2x​)

$$\underset{x\rightarrow 0}{\mathrm{lim}}2\times (\frac{{x}^{2}-{\mathrm{sin}}^{2}x}{4{x}^{4}})(\frac{0}{0}\text{form})(\because \underset{t\rightarrow 0}{\mathrm{lim}}\frac{\mathrm{sin}t}{t}=1)$$x→0lim​2×(4x4x2−sin2x​)(00​form)(∵t→0lim​tsint​=1)

Applying L'Hospital's Rule,

$$\underset{x\rightarrow 0}{\mathrm{lim}}2\times (\frac{2x-2\mathrm{sin}x\mathrm{cos}x}{4\cdot 4{x}^{3}})=\underset{x\rightarrow 0}{\mathrm{lim}}(\frac{2x-\mathrm{sin}2x}{8{x}^{3}})(\frac{0}{0}\text{form})$$x→0lim​2×(4⋅4x32x−2sinxcosx​)=x→0lim​(8x32x−sin2x​)(00​form)

$$\underset{x\rightarrow 0}{\mathrm{lim}}(\frac{2-2\mathrm{cos}2x}{24{x}^{2}})(\frac{0}{0}\text{form})$$x→0lim​(24x22−2cos2x​)(00​form)

$$\underset{x\rightarrow 0}{\mathrm{lim}}(\frac{4\mathrm{sin}2x}{48x})=\underset{x\rightarrow 0}{\frac{1}{6}\mathrm{lim}}(\frac{\mathrm{sin}2x}{2x})=\frac{1}{6}$$x→0lim​(48x4sin2x​)=x→061​lim​(2xsin2x​)=61​