Question

The value of $b$; if $\underset{x\to 0}{lim}\left[\frac{a{e}^x-b\cos x+c{e}^{-x}}{x\sin x}\right]=2$; is

• A. $-2$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

The correct option is D $2$

$\underset{x\to 0}{lim}\frac{a(1+x+\frac{x^2}{2!}+...)-b(1-\frac{x^2}{2}+...)+c(1-x+\frac{x^2}{2!})}{x^2}\times \underset{x\to 0}{lim}\frac{x}{\sin x}=2$

$\Rightarrow \underset{x\to 0}{lim}\frac{a-b+c}{x^2}+\frac{a-c}{x}+(\frac{a}{2}+\frac{b}{2}+\frac{c}{2})+...=2$

$LHS$ can be equal to $a$ finite quantity only if $a-b+c=0$ ...(1)

and $a-c=0$ ...(2)

When (1) and (2) are satisfied then $LHS$ is equal to $2$ only if

$\frac{a+b+c}{2}=2$ ...(3)

Solving (1), (2) and (3)

$b=2$