Solve without L'Hopital's rule: $\lim_{x\to0}\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}$ Solve without L'Hopital's rule:

$$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}$$

My work:
$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}=\displaystyle\lim_{x\to0}{\frac{\cos{(2x)}\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\sin{(2x)}}}$
$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\sin{2x}}}=\displaystyle\lim_{x\to0}{\frac{\cosh(3x^2)\cdot e^{8x^3}-1}{x^2\sin{(2x)}}}\cdot\displaystyle\lim_{x\to0}{\frac{1}{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}+1}}$
All of my attempts failed.
$$\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}+1\;\;\text{is continuous &decreasing }$$
Source: Matematička analiza 1, 2. kolokvij

Using Taylor series from the strating point.
Use the standard series of $\cosh(t)$ and make $t=3x^2$ gives
$$\cosh(x^3)=1+\frac{9 x^4}{2}+O\left(x^{8}\right)$$ Using the binomial expansion or Taylor series again
$$\sqrt{\cosh(x^3) }=1+\frac{9 x^4}{4}+O\left(x^{8}\right)$$
Use the standard series of $e^t$ and make $t=4x^3$ gives
$$e^{4x^3}=1+4 x^3+O\left(x^6\right)$$
Use the standard series of $\tan(t)$ and make $t=2x$ gives
$$\tan(2x)=2 x+\frac{8 x^3}{3}+O\left(x^5\right)$$
Combining everything
$$\frac{\sqrt{\cosh{(3x^2)}}\, e^{4x^3}-1}{x^2\tan(2x)}=\frac {4 x^3+\frac{9 x^4}{4}+O\left(x^6\right) } {2 x^3+\frac{8 x^5}{3}+O\left(x^7\right) }$$ Now, long division to get
$$\frac{\sqrt{\cosh{(3x^2)}}\, e^{4x^3}-1}{x^2\tan(2x)}=2+\frac{9 x}{8}-\frac{8 x^2}{3}+O\left(x^3\right)$$
Use your pocket calculator with $x=0.1$; you should get $2.08865$ while the above truncated expansion gives $2.08583$.
