Question

Find the values of $a,b$ & $c$, so that $\underset{x\to 0}{lim}\frac{a{e}^x-b\cos x+c{e}^{-x}}{x.\sin x}=2$.
Find $a+b+c$?

Answer 1

$\underset{x\to 0}{lim}\frac{a{e}^x-b\cos x+c{e}^{-x}}{x\sin x}=2$
$\underset{x\to 0}{lim}\frac{a{e}^x+c{e}^{-x}-b\cos x}{x^2}=2$
$\underset{x\to 0}{lim}\frac{a[1+x+\frac{x^2}{2!}+...]+c[1-x+\frac{x^2}{2!}+...]-b+b-b\cos x}{x^2}$ $=2$
$\Rightarrow a+c-b=0,a-c=0$ & $\frac{a+c}{2}+\frac{b}{2}=2$
$\Rightarrow a=c=1,b=2$