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

We observe that as $x\to 0$, numerator tends to $a -b + c$ whereas the denominator tends to 0. Therefore, for the limit to exist we must have 

$a-b+c=0$               …(i)

Now, if $a-b+c=0,$ then

$\underset{x\to 0}{lim} \frac{a{e}^x-b\cos x+c{e}^{-y}}{x\sin x}$is in the $\frac{0}{0}$ form.

$\therefore \underset{x\to 0}{lim} \frac{a{e}^x+b\sin x-c{e}^{-x}}{\sin x+x\cos x}=2$    [Using L' Hospital's Rule] 

Here, the numerator is tending to $a-c$ as $x\to 0$. Therefore for 

the limit to exist, we must have 

$a-c=0$                                    …(ii)

$\underset{x\to 0}{lim} \frac{a{e}^x+b\sin x-c{e}^{-x}}{\sin x+x\cos x}$ is in the form $\frac{0}{0}$ 

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\underset{x\to 0}{lim} \frac{a{e}^x+b\cos x+c{e}^{-x}}{2\cos x-x\sin x}=2\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{a+b+c}{2}=2\Rightarrow a+b+c=4\end{array}$