Finding The Lim Of Expression With Trigonometry 
Evaluate
  $$\lim_{n\rightarrow \infty}\left(\frac{\sin ^3n^2-5\cdot \sin ^2n^2+3}{\sqrt{\ln \left( \left| \dfrac{1}{\tan ^{19}e^ {- n }}+3\right| \right)+\sin \left( \dfrac{1}{\tan ^{24}e^ {- n }} \right)}}\right)$$

I have found out that when $n\rightarrow \infty$ limit = $0$, but I am sure that there are some properties of $\ln$, $\sin$, $\tan$ that can make it much easier.
Does $\sin$ and $\cos$ limits are $[-1,1]$ and $\tan$ $[0,\infty]$ ?
What about $\ln$ ?

Value of
$$\sin^3n^2-5\sin^2n^2+3$$
for all $n\in\mathbb{R}$ must be in interval $[-3,3]$.
Now you can see that
$$\lim_{n\to\infty}\ln\left|\dfrac1{\tan^{19}e^{-n}}+3\right|=\ln\left|\dfrac1{\tan^{19}\lim_{n\to\infty}e^{-n}}+3\right|=\infty$$
Finally
$$\sin\dfrac1{\tan^{24}e^{-n}}$$
must be in interval $[-1,1]$.
Now you have two bounded intervals and function that tends to $\infty$, so you can write limit as
$$\sqrt{\lim_{n\to\infty}\dfrac{c_1}{n+c_2}}$$
where $c_1$ and $c_2$ are some constants in bounded interval. You can clearly see that answer is $0$.
