Given that

$$\lim_{x\rightarrow0}\frac{ae^x - be^{-x}}{x + \sin x} = 1$$

Then the value of ab is

Since the expression $$\lim_{x\rightarrow0}\frac{ae^x - be^{-x}}{x + \sin x}$$ is in 0/0 form, as RHS is a finite number, and denominator is zero, hence, the numerator must also be zero to get a finite number. Hence, a-b=0 when x tends to 0.

L'Hospitals rule can be applied 

$$\lim_{x \rightarrow 0}\frac{ae^x + be^{-x}}{1 + \cos x}$$ = 1

$$\frac{a+b}{2}$$ = 1 which is valid only when a = b = 1

$$\therefore$$ ab=1

Hence D is the correct answer.