Question

If $\underset{x\to 0}{lim}\frac{a{e}^x-b\cos x+c{e}^{-x}}{x\sin x}=2,$, then $a+b+c$ is equal to

Answer 1

$\underset{x\to 0}{lim}\frac{\{a(1+x+\frac{x^2}{2!}+\cdot )-b(1-\frac{x^2}{2!}+\frac{x^4}{4!}\cdots )+c(1-x+\frac{x^2}{2!})\}}{x(x-\frac{x^3}{3!}+\cdots )}=2\therefore \underset{x\to 0}{lim}\frac{(a-b+c)+x(a-c)+x^2(\frac{a}{2}+\frac{b}{2}+\frac{c}{2})+\cdots }{x^2(1-\frac{x^2}{6}\cdots )}\therefore a-b+c=0\&a-c=0\&\frac{a}{2}+\frac{b}{2}+\frac{c}{2}=2\Rightarrow a+b+c=4$