<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] You need to make use of the following standard limits $$\lim_{t\to 0}\frac{e^t-1}{t}=1,\lim_{t\to 0}\frac{\tan t} {t} =1,\lim_{t\to 0}\frac{\sinh t} {t} =1,\lim_{t\to a} \frac{t^n-a^n} {t-a} =na^{n-1}$$ Using the second limit above the limit in question is equal to the limit of $$\frac{\sqrt{\cosh(3x^2)}e^{4x^3}-1}{2x^3}$$ Adding and subtracting $e^{4x^3}$ in numerator we can split the above fraction as $$e^{4x^3}\cdot\frac{\cosh^{1/2}(3x^2)-1}{2x^3}+\frac{e^{4x^3}-1}{2x^3}$$ and this can be further rewritten as $$e^{4x^3}\cdot\frac{\cosh^{1/2}(3x^2)-1}{\cosh(3x^2)-1}\cdot\frac{\cosh(3x^2)-1}{2x^3}+2\cdot\frac{e^{4x^3}-1}{4x^3}$$ The limit of above expression is equal to that of $$1\cdot\frac{1}{2}\cdot\frac{\cosh(3x^2)-1}{2x^3}+2$$ via the standard limits mentioned at the start of the answer. Next we can use the identity $$\cosh t =1+2\sinh^2(t/2)$$ to rewrite the first fraction as $$\frac{\sinh^2(3x^2/2)}{(3x^2/2)^2}\cdot \frac{(3x^2/2)^2}{2x^3}$$ which tends to $1^2\cdot 0=0$. The desired limit is thus $2$.

A combination of algebraic manipulation and standard limits is sufficient to solve most limit problems encountered in a typical calculus course.
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