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
Best Answer
$${\sqrt{\cosh(3x^2)}e^{4x^3}-1\over x^2\tan(2x)}={1\over\sqrt{\cosh(3x^2)}e^{4x^3}+1}\cdot{x\over\tan(2x)}\cdot{\cosh(3x^2)e^{8x^3}-1\over x^3}$$
and
$${\cosh(3x^2)e^{8x^3}-1\over x^3}={e^{8x^3}-1\over x^3}+e^{8x^3}{\cosh(3x^2)-1\over x^3}={e^{8x^3}-1\over x^3}+{e^{8x^3}\over\cosh(3x^2)+1}\left(\sinh(3x^2)\over x^2\right)^2x$$
Now it's easy to see that
$${1\over\sqrt{\cosh(3x^2)}e^{4x^3}+1}\to{1\over1+1}={1\over2}$$
$${x\over\tan(2x)}={\cos x\over2}\cdot{2x\over\sin(2x)}\to{1\over2}\cdot1={1\over2}$$
$${e^{8x^3}-1\over x^3}=8\cdot{e^{8x^3}-1\over8x^3}\to8\cdot1=8$$
and
$${e^{8x^3}\over\cosh(3x^2)+1}\left(\sinh(3x^2)\over x^2\right)^2x={e^{8x^3}\over\cosh(3x^2)+1}\left(\sinh(3x^2)\over3x^2\right)^2(9x)\to{1\over1+1}\cdot1^2\cdot0=0$$
and thus
$${\sqrt{\cosh(3x^2)}e^{4x^3}-1\over x^2\tan(2x)}\to{1\over2}\cdot{1\over2}(8+0)=2$$