The limit

The given function can be re-written as:

$\displaystyle \dfrac{ ae^x + ce^{-x} - b \dfrac{ \log(1+x)} { x} } { x^2 \dfrac{ \sin x } {x}}$

Note that $\log (1+x) = x - \dfrac {x^2} 2 + \dfrac{x^3} 3 + O( x^4 )$ for $|x| < 1$ .

Also, $e^x = 1 + x + \dfrac{x^2} {2!} + O(x^3)$

Substituting these expansions in the given function, we get:

$\displaystyle \dfrac{ a\left( 1 + x + \dfrac{x^2} {2!} + O(x^3) \right) + c\left( 1 - x + \dfrac{x^2} {2!} + O(x^3) \right) - b\left( 1 - \dfrac x 2 + \dfrac{x^2} 3 + O(x^3) \right) } { x^2 \dfrac{ \sin x } {x}}$

Rearranging the terms slightly, we get:

$\displaystyle \dfrac{ (a + c - b) + x(a - c + \dfrac b 2) + x^2 \left( \dfrac {a+c} 2 - \dfrac b 3 \right) + O(x^3) } { x^2 \dfrac{ \sin x } {x}}$

Now, for this limit to exist, we must have $a+c-b = 0$ and $a - c + \dfrac b 2 = 0$ . Also, for the limit to be $2$ , this too should hold good: $\dfrac {a+c} 2 - \dfrac b 3 = 2$ .

Solving the three equations, we get $a + b + c = 24$ .