The value of $b$, if $\lim_{x \to 0}(\frac{ae^x-b\cos x+ce^{-x}}{x\sin x})=2$ Question : 
The value of $b$, if 
$$\lim_{x \to 0^+}(\frac{ae^x-b\cos x+ce^{-x}}{x\sin x})=2$$
Solution : $$= \lim_{x \to 0^+} \frac{\log_e\left[\frac{4^x-3^x}{4^x-1}\right]}{x}(4^x-1)$$
$$= \lim_{x \to 0^+} \frac{\left[1-\frac{3^x-1}{4^x-1}\right]}{\left(\frac{3^x-1}{4^x-1}\right)}\times  \frac{3^x-1}{4^x-1}\times  \frac{4^x-1}{x}$$
I am unable to take it further. Please help on this thanks

Using Taylor series and taking an asymptotic equivalent of the order $2$ we get
$$\frac{ae^x-bcosx+ce^{-x}}{x\sin x}\sim_0\frac{a+ax+a\frac{x^2}2-b+b\frac{x^2}{2}+c-cx+c\frac{x^2}{2}}{x^2}$$
so we see that the desired limit is $2$ if and only if
$$a-b+c=0\quad;\quad a-c=0\quad;\quad a+b+c=4$$
hence we get $b=2$.
