JEE Main 2021 — Mathematics Calculus

Given $\underset{x\rightarrow 0}{\mathrm{lim}}\frac{a{e}^{x}-b\mathrm{cos}x+c{e}^{-x}}{x\mathrm{sin}x}=2$x→0lim​xsinxaex−bcosx+ce−x​=2

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}\frac{a(1+x+\frac{{x}^{2}}{2!}\ldots )-b(1-\frac{{x}^{2}}{2!}+\ldots )+c(1-x+\frac{{x}^{2}}{2!})}{(\frac{x\mathrm{sin}x}{x})x}=2$$⇒x→0lim​(xxsinx​)xa(1+x+2!x2​…)−b(1−2!x2​+…)+c(1−x+2!x2​)​=2

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}\frac{a(1+x+\frac{{x}^{2}}{2!}\ldots )-b(1-\frac{{x}^{2}}{2!}+\ldots )+c(1-x+\frac{{x}^{2}}{2!})}{{x}^{2}}=2(\because \underset{x\rightarrow 0}{\mathrm{lim}}\frac{\mathrm{sin}x}{x}=1)$$⇒x→0lim​x2a(1+x+2!x2​…)−b(1−2!x2​+…)+c(1−x+2!x2​)​=2(∵x→0lim​xsinx​=1)

For the given limit to exist numerator should be $0,\text{when }x\rightarrow 0.$0,when x→0.

Then, $a-b+c=0$a−b+c=0 $...(i)$...(i)

Applying L'Hospital's rule, we get

$$\underset{x\rightarrow 0}{\mathrm{lim}}\frac{a(1+\frac{2x}{2!}\ldots )-b(-\frac{2x}{2!}+\ldots )+c(-1+\frac{2x}{2!})}{2x}=2$$x→0lim​2xa(1+2!2x​…)−b(−2!2x​+…)+c(−1+2!2x​)​=2

Again, for the given limit to exist numerator should be $0,\text{when}x\rightarrow 0.$0,whenx→0.

Then, $a-c=0...(\mathrm{ii})$a−c=0...(ii)

Applying, L'Hospital's rule, we get

$$\underset{x\rightarrow 0}{\mathrm{lim}}\frac{a(1+\ldots )-b(-1+\ldots )+c(1)}{2}=2$$x→0lim​2a(1+…)−b(−1+…)+c(1)​=2

$$\Rightarrow \frac{a+b+c}{2}=2$$⇒2a+b+c​=2

$$\Rightarrow a+b+c=4$$⇒a+b+c=4