$As\: x \to 0, x - \sin x \to 0 \ \text{The limit exists and the denominator approaches 0, } \ \text{Hence, for a }\dfrac{0}{0} \text{ form limit, the numerator must also approach 0}$ $ \ \therefore, \lim \limits_{x \to 0} ae^x - be^{-x} +cx = 0 \ a -b+0 = 0 \ \implies a =b ...(A)$ $

\ \lim \limits_{x \to 0}\dfrac{ ae^x - be^{-x} +cx}{x -\sin x} = 4 \ \text{Using L'Hopital's rule,}$ $ \ \lim \limits_{x \to 0}\dfrac{ ae^x + be^{-x} +c}{1 -\cos x} = 4 \ As \: x \to 0, 1-\cos x \to 0$ $ \ \text{The limit exists and the denominator approaches 0, the numerator must also approach 0.} \ \lim \limits_{x\to 0} ae^x +be^{-x} +c =0 $ $\ a+b+c = 0, 2a+c =0 ... from (A) \ c = -2a ....(B) $ $\ \text{Using L'Hopital's rule again,} \ \lim \limits_{x\to0} \dfrac{ae^x - be^{-x}}{\sin x} = 4$ $\ \text{Using L'Hopital's rule again,} \ \lim \limits_{x\to0} \dfrac{ae^x +be^{-x}}{\cos x} = 4$ $\dfrac{ae^0 +be^0 }{\cos 0} =4 \ a+b = 4 $ $ \ a= b= 2 ...from (A) \ c= -2(2) = -4 ...from (B) \ \bbox[3pt, yellow]{a=2,b=2,c=-4}$