Help with why $\lim_{n\to\infty}\frac{13n^3+2n^2+6n\log(n)}{n^3}=13$ Reposting this since I apparently posted in the wrong website. 
Anyway, I just want to know, step by step, how did the guy reach 13 here:
$$\begin{align*}
\lim_{n\to\infty}\frac{T(n)}{f(n)} &= \lim_{n\to\infty}\frac{13n^3+2n^2+6n\log(n)}{n^3}\\
&= \lim_{n\to\infty}\left(13+\frac{2}{n}+\frac{6\log(n)}{n^2}\right)\\
&=13
\end{align*}$$
(original image)
That's all. I have googled and clicked every link, no real answer.
I'm not even studying limits, I'm studying algorithm complexity, but I need to know step by step how that result was achieved. Any help is appreciated.
 A: Divide everything by $n^3$ in the first step.
So $\displaystyle\frac{2}{n} \to 0$ and $\displaystyle\frac{6\log n}{n^2} \to 0 \text{ as } n \to \infty$ because $n$ is very large. Thus you get $13$.
A: We may use the definition of the limit directly here as follows.  From the triangle inequality, along with the inequality $\log n<n$, we have for any given $\epsilon>0$
$$\begin{align}
\left|\left(13+\frac2n +\frac{6\log n}{n^2}\right)-13\right|&\le \left|\frac2n\right|+\left|\frac{6\log n}{n^2}\right|\\\\
&\le \frac8n\\\\
&< \epsilon
\end{align}$$
whenever $n\ge N=\lceil \frac{8}{\epsilon}\rceil$.  And we are done!
A: Let  $$\displaystyle L = \lim_{n\rightarrow \infty}\frac{13n^3+2n^2+6n\cdot \ln n}{n^3} = \lim_{n\rightarrow \infty}\frac{13n^3}{n^3}+\lim_{n\rightarrow \infty}\frac{2n^2}{n^3}+\lim_{n\rightarrow \infty}\frac{6n\cdot \ln n}{n^3}$$
So we get Limit $$\displaystyle L = 13+\lim_{n\rightarrow \infty}\frac{2}{n}+\lim_{n\rightarrow \infty}\frac{\ln n}{n^2} = 13+0+\underbrace{\lim_{n\rightarrow \infty}\frac{\ln n}{n^2}}_{J}$$
So For calculation of $$\displaystyle J = \lim_{n\rightarrow \infty}\frac{\ln n}{n^2}$$
Using $\bf{D,LHopital\; Rule}$
So $$\displaystyle J = \lim_{n\rightarrow \infty}\frac{1}{n\cdot 2n}=\lim_{n\rightarrow \infty}\frac{1}{2n^2} = 0$$
So we get Limit $$L = 13+0+0 = 13$$
A: Notice, we have $$\lim_{n\to \infty}\frac{13n^3+2n^2+6n\log (n)}{n^3}$$
Using L-Hospital's rule as follows $$=\lim_{n\to \infty}\frac{39n^2+4n+6\log (n)+6}{3n^2}$$
$$=\lim_{n\to \infty}\frac{78n+4+\frac{6}{n}}{6n}$$ $$=\lim_{n\to \infty}\frac{78-\frac{6}{n^2}}{6}$$ $$=\lim_{n\to \infty}\frac{78}{6}-\lim_{n\to \infty}\frac{1}{n^2}$$ $$=13-0=13$$
A: To remove indetermenancy in case of infinity by infinite form we will have to divide the variable in numerator with denominators varible's highest exponent
