Taylor's Expansion Exercise I have a few troubles coming up with the solution of this limit.
$$ \lim_{x \to 0} \frac {8x^2(e^{6x}-1)}{2x-\sin(2x)}$$
I've tried using Taylor like this but I honestly have no idea if it's even remotely close
$$\lim_{x\to 0} \frac{8x^2(1+6x+o(x)-1)}{2x-2x+o(x)}$$
Thank you so much
 A: Using
$$ e^{ax} - 1 = (a x) \, \left(1 + \frac{a x}{2} + \frac{a^2 x^2}{6} + \mathcal{O}(x^3) \right) $$
and
$$ b x - \sin(b x) = \frac{b^3 x^3}{3!} \ \left(1 - \frac{b^2 x^2}{20} + \mathcal{O}(x^4) \right) $$
then
\begin{align}
\frac{x \, \left(e^{a x} - 1\right)}{b x - \sin(b x)} &= \frac{ a x^3 \, \left(1 + \frac{a x}{2} + \frac{a^2 x^2}{6} + \mathcal{O}(x^3) \right) }{\frac{b^3 x^3}{6} \, \left(1 - \frac{b^2 x^2}{20} + \mathcal{O}(x^4) \right) } \\
&= \frac{6 a}{b^3} \, \left( 1 + \frac{a x}{2} + \left(\frac{b^2}{20} + \frac{a^2}{6}\right) \, x^2 + \mathcal{O}(x^3) \right).
\end{align}
The limit now takes the form
\begin{align}
\lim_{x \to 0} \, \frac{\alpha x \, \left(e^{a x} - 1\right)}{b x - \sin(b x)} &= \lim_{x \to 0} \, \frac{6 \alpha a}{b^3} \, \left( 1 + \frac{a x}{2} + \left(\frac{b^2}{20} + \frac{a^2}{6}\right) \, x^2 + \mathcal{O}(x^3) \right) \\
&= \frac{6 \alpha a}{b^3}.
\end{align}
The proposed limit has $\alpha = 8$, $a = 6$, and $b = 2$, which gives
$$ \lim_{x \to 0} \, \frac{8 x \, \left(e^{6 x} - 1\right)}{2 x - \sin(2 x)} = 36. $$
A: $$ \lim_{x \to 0} \dfrac{8x^2(e^{6x}-1)}{2x-\sin 2x}$$
$$\dfrac{8x^2(e^{6x}-1)}{2x-\sin 2x}\sim \dfrac{8x^2(1+6x-1)}{2x-2x+\dfrac{4x^3}{3}}\to 36 \quad \text{as} \quad x\to 0$$
