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

We start with Taylorexpanding the numerator and denominator:
\begin{align}
&\lim_{x \to 0^+}\left(\frac{ae^x-b\cos(x)+ce^{-x}}{x\sin(x)}\right)\\
=&\lim_{x \to 0^+}\left(\frac{a+ax+\frac{1}{2}ax^2-b(1-\frac{1}{2}x^2)+c-cx+\frac{1}{2}cx^2+O(x^3)}{x^2+O(x^3)}\right)\\
=&\lim_{x \to 0^+}\left(\frac{a+c-b+(a-c)x+\frac{1}{2}(a+b+c)x^2+O(x^3)}{x^2+O(x^3)}\right)=2\\
\end{align}
From this follows $a+c-b=0$ and $a+b+c=4$. We conclude that $b=2$
