Seems so complex !

i did this sum obviously ignoring the typo of $1+3n$ instead of $1+3x$

---

We have a horrendous looking expression to ourselves.. but this is going to be solved by making it more horrendous looking:

our expression:

$\large{\lim\limits_{x\to 0}\dfrac{tan\sqrt[3]{x}.ln(1+3x)}{(tan^{-1}\sqrt{x})^{2}.(e^{5\sqrt[3]{x}}-1)}}$ ........... $(i)$

we take: $tan^{-1}\sqrt{x}=z \implies x=tan^{2}z$

so that: $x\to 0 \implies z\to 0$

we also write: $\large{\dfrac{ln(1+3x)}{(tan^{-1}\sqrt{x})^{2}}=\dfrac{ln(1+3.tan^{2}z)}{3.tan^{2}z}.\dfrac{tan^{2}z}{z^{2}}.3}$

now we write the horrendous expression into another more horrible looking (but beautiful) expression:

$\large{\lim\limits_{x\to 0}\frac{1}{5}.(\dfrac{tan\sqrt[3]{x}}{\sqrt[3]x}).(\dfrac{1}{\frac{e^{5\sqrt[3]{x}}-1}{5\sqrt[3]{x}}}).(\dfrac{ln(1+3.tan^{2}z)}{3.tan^{2}z}).(\dfrac{tan^{2}z}{z^{2}}).3}$

Note that. applying limits in each of the brackets, we get $1$

so, our end result: $\large{\frac{1}{5}.1.1.1.1.3=\frac{3}{5}=\boxed{0.6}}$

---

@Abhishek Singh what do you think? :D

(1+3n)===》》(1+3x)

Please do correct the term inside ln(1+3n) to ln(1+3x)