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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 $\ln$ ?

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    $\begingroup$ Ummm where are you taking your limit to? $\endgroup$ – EhBabay Dec 3 '14 at 20:48
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    $\begingroup$ You want the limit at infinity or at 0? $\endgroup$ – PhzksStdnt Dec 3 '14 at 20:49
  • $\begingroup$ ahh sorry, yes when $n\rightarrow \infty$. I will edit the question $\endgroup$ – gbox Dec 3 '14 at 20:52
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    $\begingroup$ @gbox Note that the numerator oscillates between finite values but in the denominator, the $e^{-n}$ term will cause the argument of $\ln$ to tend to $\infty$ and thus we can easily say that the limit = $0$ $\endgroup$ – MathGod Dec 3 '14 at 21:15
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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 $0$.

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  • $\begingroup$ Why do the interval is [-3,4], why 4? $\endgroup$ – gbox Dec 3 '14 at 23:40
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    $\begingroup$ Yes, it is in interval $[-3,3]$. I edited answer. $\endgroup$ – user164524 Dec 9 '14 at 16:02

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