JEE Main 2024 — Mathematics Calculus

Given, $\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{e}^{2|\mathrm{sin}x|}-2|\mathrm{sin}x|-1}{{x}^{2}}$x→0lim​x2e2∣sinx∣−2∣sinx∣−1​

Now, using the expansion of ${e}^{x}=1+\frac{x}{1!}+\frac{{x}^{2}}{2!}+........\infty$ex=1+1!x​+2!x2​+........∞ we get,

$$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{(1+\frac{|2\mathrm{sin}x|}{1!}+\frac{{|2\mathrm{sin}x|}^{2}}{2!}+.....\infty )-|2\mathrm{sin}x|-1}{{x}^{2}}$$=x→0lim​x2(1+1!∣2sinx∣​+2!∣2sinx∣2​+.....∞)−∣2sinx∣−1​

$$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\frac{{|2\mathrm{sin}x|}^{2}}{2!}+\frac{{|2\mathrm{sin}x|}^{3}}{3!}.....\infty }{{x}^{2}}$$=x→0lim​x22!∣2sinx∣2​+3!∣2sinx∣3​.....∞​

$$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{2{\mathrm{sin}}^{2}x+\frac{{|2\mathrm{sin}x|}^{3}}{3!}.....\infty }{{x}^{2}}$$=x→0lim​x22sin2x+3!∣2sinx∣3​.....∞​

$$=2{\text{as}\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{\mathrm{sin}}^{2}x}{{x}^{2}}=1\text{ and other terms will become zero}}$$=2asx→0lim​x2sin2x​=1 and other terms will become zero