Evaluating $\lim_{x\to 0}\frac{1-\cos(x^2)}{x^3(4^x-1)}$ 
I need to calculate
$$\lim_{x\rightarrow 0}\frac{1-\cos(x^2)}{x^3(4^x-1)}$$
and the options are:
(a) $\frac 12 \ln 2\quad$ (b) $\ln 2\quad$ (c) $\ln 4\quad$ (d) $1 - \frac 12 \ln \left( \frac{e^2}{4}\right)$.

The answers would are given to be $b$ and $d$
I tried to solve it in the following manner:
\begin{align}\lim_{x\rightarrow 0}\frac{1-\cos(x^2)}{x^3(4^x-1)} &=\lim_{x\rightarrow 0}(\frac{2\sin^2(\frac{x^2}{2})}{x^4}\cdot\frac{x}{4^x-1})\\
&=\lim_{x\rightarrow 0}(\frac{2\sin^2(\frac{x^2}{2})}{(\frac{x^2}{2})^2\cdot 4}\cdot\frac{x}{4^x-1})\\
&=\lim_{x\rightarrow 0}(\frac{1}{2}\frac{\sin^2(\frac{x^2}{2})}{(\frac{x^2}{2})^2})\lim_{x\rightarrow 0}(\frac{x}{4^x-1})\\
&=\frac{1}{2}\frac{1}{\ln(4)}\\
&=\frac{1}{4}\log_2(e).
\end{align}
Is my solution correct? Or am I missing something?

As @Joe said, it would be simpler to write it as $\frac 1{2 \ln (4)}$, but there is nothing wrong with your solution.
