<s>[INST] 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
 [/INST] \begin{align*}
\dfrac{\sqrt{\cosh(3x^{2})}e^{4x^{3}}-1}{x^{2}\tan(2x)}&=\dfrac{\cosh(3x^{2})e^{8x^{3}}-1}{x^{2}(2x)}\dfrac{2x}{\tan(2x)}\dfrac{1}{\sqrt{\cosh(3x^{2})}e^{4x^{3}}+1},
\end{align*}
so the issue is to compute
\begin{align*}
\lim_{x\rightarrow 0}\dfrac{\cosh(3x^{2})e^{8x^{3}}-1}{x^{2}(2x)}.
\end{align*}
Note that
\begin{align*}
\dfrac{\cosh(3x^{2})e^{8x^{3}}-1}{x^{2}(2x)}=\dfrac{(e^{3x^{2}}+e^{-3x^{2}})e^{8x^{3}}-2}{2(2x^{3})}.
\end{align*}
If you are allowed to use power series, then 
\begin{align*}
(e^{3x^{2}}+e^{-3x^{2}})e^{8x^{3}}-2\approx 16x^{3},
\end{align*}
so the cancellation with the $x^{3}$ in the denominator is safe.
</s>