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?

binomial theorem:
$(a+b)^n) = a^n + na^{n-1}b + \frac {n(n-1)}{2} a^{n-1}b^2 \cdots$
works for fractional exponents
$(1+4x)^\frac 12 = 1 + \frac 12 (4x) - \frac 18 (4x)^2+ \cdots%$
$\sin x = x - \frac 16 x^2+\cdots\\
\sin 2x = (2x) - \frac16 (2x)^3+ \cdots$
numerator: $1 + 2x - 2x^2 - 1 - 2x + \frac 86 x^3 = -2x^2- \frac 16 x^3 + \cdots$
denominator:
$\ln (1 + x) = x - \frac 12 x^2 + \frac 13 x^3\\
\ln(1+x^2) = x^2 - \frac 12 x^4\cdots$  
$-2$
