Limit of a function tending to a finite number If 
$$\lim_{x\to 0} \frac{ae^x - b\cos x +ce^{-x}}{x\sin x} = 2$$
then find the value of $a+b+c$.
My book has given the following solution to the above problem :- 
We observe that as $x$ tends to zero , numerator tends to $a-b+c$ whereas the denominator tends to zero. Therefore for the limit to exist , we must have ,$a-b+c=0$
Now I am really confused at this point. Why would we want the numerator to attain the value of $0$ . Wouldn’t that give us an indeterminate answer? But actually it’s suposed to be two . Can you please explain ? Thank you for your help. 
 A: Hint. By using Taylor expansions at $0$,
$$\frac{ae^x - b\cos x +ce^{-x}}{x\sin x} =\frac{a(1+x+\frac{x^2}{2}) - b(1-\frac{x^2}{2}) +c(1-x+\frac{x^2}{2})+o(x^2)}{x(x+o(x))}\\
=\frac{(a-b+c)+(a-c)x+\frac{(a+b+c)}{2}x^2+o(x^2)}{x^2+o(x^2)}.$$
In order to have the final limit $2$ we need that
$$(a-b+c)=0,\quad(a-c)=0,\quad \frac{(a+b+c)}{2}=2.$$
Can you take it from here?
P.S. If you prefer to use L'Hopital begin with your initial observation
$$\lim_{x\to 0}ae^x - b\cos x +ce^{-x}=a-b+c.$$
If $a-b+c=0$ then, by L'Hopital,
$$\lim_{x\to 0}\frac{ae^x - b\cos x +ce^{-x}}{x}=\lim_{x\to 0}\frac{ae^x + b\sin x -ce^{-x}}{1}=a-c.$$
If $a-b+c=0$ AND $a-c=0$ then, by L'Hopital (twice),
$$\begin{align}
\lim_{x\to 0}\frac{ae^x - b\cos x +ce^{-x}}{x^2}
&=\lim_{x\to 0}\frac{ae^x + b\sin x -ce^{-x}}{2x}
\\&=\lim_{x\to 0}\frac{ae^x + b\cos x +ce^{-x}}{2}=\frac{a+b+c}{2}.
\end{align}$$
A: If $a-b+c \neq 0$, the limit would be either $+\infty$ or $-\infty$. 
Note, as written as Robert Z, that
$$
\frac{ae^x-b\cos x+cx^{-x}}{x\sin x}=\frac{(a-b+c)+x(a-c)+x^2(\frac{a}{2}+\frac{b}{2}+\frac{c}{2}))+O(x^3)}{x^2+O(x^3)}
$$
as $x\to 0$. Since the limit exists, you need $a-b+c=0$, $a=c$ and $\frac{a}{2}+\frac{b}{2}+\frac{c}{2}=2$. Therefore $$a+b+c=4$$ (and you also know that $a=1, b=2, c=1$).
A: If $a - b + c \ne 0$ then
$$\frac{ae^x-b\cos x+ce^{-x}}{x\sin x} \xrightarrow{x\to 0} \frac{a-b+c}{0} = \operatorname{sgn}(a-b+c) \cdot \infty \ne 2$$
Therefore $a - b + c = 0$. The limit is now an indeterminate form $\frac00$ so we can apply L'Hopital once. Differentiating gives
$$\frac{ae^x+b\sin x-ce^{-x}}{x\cos x+\sin x}$$
Again, if $a+b-c \ne 0$ then we would have:
$$\frac{ae^x+b\sin x-ce^{-x}}{x\cos x+\sin x} \xrightarrow{x\to 0} \frac{a+b-c}{0} = \operatorname{sgn}(a+b-c) \cdot \infty \ne 2$$
Therefore $a+b-c = 0$. The limit is now an indeterminate form $\frac00$ so we can again apply L'Hopital. Differentiating gives
$$\frac{ae^x+b\cos x+ce^{-x}}{-x\sin x+2\cos x}$$
Now the denominator is $\ne 0$ when $x\to 0$ so:
$$\frac{ae^x+b\cos x+ce^{-x}}{-x\sin x+2\cos x}\xrightarrow{x\to 0} \frac{a+b+c}{2}$$
This limit has to be $2$ so $$a+b+c = 4$$
$$a-b+c = 0$$
$$a+b-c = 0$$
The only solution is $a = 0, b = 2, c = 2$.
