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Question:

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

Solution:

$\lim _{x \rightarrow 0} \frac{a e^{x}-b_{\cos x+c e^{-x}}}{x \sin x}=2$

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

$a-b+c=0$

$a-c=0$

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

$\Rightarrow a+b+c=4$