<s>[INST] 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.
 [/INST] Since you have tagged this with "taylor-expansion", you might want to see your limit as 
$$\lim_{x\to 0} \cfrac{a(1+x+\frac{x^2}{2}+\cdots) - b(1-\frac{x^2}{2}+\cdots) +c (1-x+\frac{x^2}{2}-\cdots) }{x(x-\frac{x^3}{6}+\cdots)}$$
Matching powers will give you $$\lim_{x\to 0} \cfrac{a-b+c}{x^2} + \cfrac{a-c}{x} + \cfrac{\frac{a}{2}+\frac{b}{2}+\frac{c}{2}}{1} + \cdots =2 $$ and so 
$$a-b+c=0$$ $$a-c=0$$ $$\frac{a}{2}+\frac{b}{2}+\frac{c}{2}=2$$
which are three equations in three unknowns, so you can solve these.
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