Evaluating limit using taylor expansions How can I evaluate this limit?  Trying with Taylor doesn't seem to give me the right result, why ?

$$\lim_{x\to 0}  \frac{\sqrt{1+4x} -1 -\sin(2x)}{\log(1+x^2)}$$

with Taylor I can approximate $\sin(2x)$ to $ 2x $
and $\log(1+x^2)$ to $ x^2 $. If I plug in these in the limits it does not give me the right limit, what am I doing wrong?

$$\log(x^2+1)\sim x^2$$
$$\sin(2x)\sim 2x -\frac{8}{6}x^3$$
$$ \sqrt{1+4x} -1\sim 2x-2x^2 $$
since $$(\sqrt{1+4x} -1)'|_{x=0} = 2~~~and ~~~(\sqrt{1+4x} -1)"|_{x=0} = -4$$
$$\lim_{x\to0}  \frac{\sqrt{1+4x} -1 -\sin{2x}}{\log{1+x^2}}\sim \lim_{x\to0}\frac{ -2x^2+\frac{8}{6}x^3}{ x^2} = -2$$
