Find the value of $a$, $b$ and $c$ for the given limit. Question - 

Find the values of $a$, $b$ and $c$ so that $$ \lim_{x\to 0} \cfrac{ae^x - b\cos x +c e^{-x} }{x\sin x} = 2 $$ 

This is what I've tried yet : 
For $ x\to 0 $ the numerator must also tend to zero as : $ e^x , \ \cos x, e^{-x} $ all $\to 1$   for $x \to 0$ . 
Therefore, we have: 
$$ 
a - b + c = 0 \\
\color{blue}{\text{OR}} \\  b = a + c \\ 
$$ 
Now, plugging this in the original equation:
$$\lim_{x\to 0} \cfrac{ae^x -(a+c)\cos x + ce^{-x} }{x^2 \times \frac{\sin x}{x} } = 2 $$
Which implies - 
$$\lim_{x\to 0} \cfrac{a(e^x - \cos x) + c(e^{-x} - \cos x) }{x^2} =2 $$
I do get a positive wave that there is some application of expansion series of $e^x \ \& \ \cos x $ . I tried that too but that just didn't work. May be I'm not doing right simplification. Any help will be greatly appreciated.

If $b\neq a+c$ the limit would not be finite. Therefore we need $b=a+c$. Using approximation $e^x=1+x+\frac{x^2}{2}+O(x^3)$, $\cos x=1-\frac{x^2}{2}+O(x^3)$, and $\sin x=x+O(x^2)$ for $x\to 0$ we get the expression
$$
\frac{a(1+x+\frac{x^2}{2})-b(1-\frac{x^2}{2})+c(1-x+\frac{x^2}{2})+O(x^3)}{x(x+O(x^2))},
$$
which simplifies [considering $b=a+c$] to
$$
\frac{(a-c)x+\frac{1}{2}(a+b+c)x^2+O(x^3)}{x^2+O(x^3)}.
$$
This implies that if $a\neq c$ then the limit would be not finite. Therefore
$$
a=c=\frac{b}{2}.
$$
In such case we would obtain that the limit is finite and it is exactly 
$$
\frac{1}{2}(a+b+c)=2a.
$$
Hence, the unique triple $(a,b,c)$ which satisfies the relation is $(1,2,1)$.
