$\lim_{n\to \infty} n^3(\frac{3}{4})^n$ I have the sequence $a_n=(n^3)\big(\frac{3}{4}\big)^n$
and I need to find whether it converges or not and the limit.
I took the common ratio which is $\frac{3 (n+1)^3}{4 n^3}$ and since $\big|\frac{3}{4}\big|<1$ it converges.
I don't know how to find the limit from here.
 A: If $\frac {a_{n+1}} {a_n} \to l$ with $|l|<1$ then $a_n \to 0$. (In fact the series $\sum a_n$ converges).
In this case $\frac {a_{n+1}} {a_n} =(1+\frac 1 n)^{3}(\frac 3 4) \to \frac 3 4$. 
A: There are a number of ways to approach this problem.  As others have mentioned, the convergence of $\displaystyle \sum^\infty n^3(3/4)^n$ implies $n^3(3/4)^n \to 0$ as $n \to \infty$, and it looks like this is what you're trying to exploit.

Exercise: If $\displaystyle \lim_{n \to \infty} a_n \neq 0$, prove that $\displaystyle \sum_{n=1}^\infty a_n$ cannot converge.

I'm a fan of logarithms myself, so here is another approach:
We have $\ln \Big( n^3(3/4)^n \Big) = 3\ln(n) + n \ln(3/4)$.  Notice that $3/4 < 1$, so its natural log will be negative; we can rearrange the expression to emphasize this:
$$3\ln(n) + n \ln(3/4) \ = \ 3 \ln(n) - n \ln(4/3)$$
At this point, we note that the linear term grows faster than the logarithmic term, so we see that $\ln(a_n) \to - \infty$, and because the natural logarithm is a continuous function, we can conclude that $a_n \to 0$.  
If you feel that the "grows faster" argument is hand-wavy, you can make it rigorous by taking a derivative, wherein you'll find that it's negative for all $x$ beyond $x \approx 10.4$ and tending to $-\ln(4/3)$, implying unbounded growth toward $- \infty$.
A: 
I can see that he proves how the sequence converges but i can't understand how he finds the limit. 

You seem to mix (the convergence of) sequence and series.

I took the common ratio which is $\frac{3}{4}$ and since $|\frac{3}{4}|<1$ it converges.

If by "it converges" you mean that the series $\sum a_n$ converges, because you seem to have applied the ratio test, then - a fortiori - you also have $a_n \to 0$, since:
$$\sum a_n \;\mbox{ converges } \implies a_n \to 0$$
