Question

$\underset{x\to 0}{lim}\frac{a{e}^x-b\cos x+c{e}^{-x}}{x\sin x}=2.$
Find the value of a,b and c.

• A. $a=0,b=1,c=1$
• B. $a=0,b=1,c=2$
• C. $a=1,b=1,c=1$
• D. $a=1,b=2,c=1$

Answer 1

Answer: (D)

$a=1,b=2,c=1$

Given that, $\underset{x\to 0}{lim}\frac{a{e}^x-b\cos x+c{e}^{-x}}{x\sin x}=2.$

Using the expansions

$e^x=1+x+\frac{x^2}{2!}+....$

$\cos x=1-\frac{x^2}{2!}+.....$

$\sin x=x-\frac{x^3}{3!}+....$

$\underset{x\to 0}{lim}\frac{a(1+x+\frac{x^2}{2!}+...)-b(1-\frac{x^2}{2!}+...)+c(1-x+\frac{x^2}{2!}-...)}{x(x-\frac{x^3}{3!}+...)}=2.$

$\Rightarrow a-b+c=0,a-c=0$ and $a+b+c=4$

Solving above equations for $a,b,c$

$\therefore a=1,b=2,c=1$

Hence, option D.