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

$\Rightarrow \underset{x\to 0}{\lim }\frac{a\left(1+x+\frac{x^2}{2!}\dots \right)-b\left(1-\frac{x^2}{2!}+\dots \right)+c\left(1-x+\frac{x^2}{2!}\right)}{\left(\frac{x\sin x}{x}\right)x}=2$

$\Rightarrow \underset{x\to 0}{\lim }\frac{a\left(1+x+\frac{x^2}{2!}\dots \right)-b\left(1-\frac{x^2}{2!}+\dots \right)+c\left(1-x+\frac{x^2}{2!}\right)}{x^2}=2 \left(\because \underset{x\to 0}{\lim }\frac{\sin x}{x}=1\right)$

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

Then, $a-b+c=0$   $...\left(\mathrm{i}\right)$

Applying L'Hospital's rule, we get 

$\underset{x\to 0}{\lim }\frac{a\left(1+\frac{2x}{2!}\dots \right)-b\left(-\frac{2x}{2!}+\dots \right)+c\left(-1+\frac{2x}{2!}\right)}{2x}=2$

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

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

Applying, L'Hospital's rule, we get 

$\underset{x\to 0}{\lim }\frac{a\left(1+\dots \right)-b\left(-1+\dots \right)+c\left(1\right)}{2}=2$

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

$\Rightarrow a+b+c=4$