specify the limit of the following equation? What is the limit of the following  $\lim_{n \to \infty}$ $\frac{(n^2-9)/(n+1)^3}{(n+3)/(3n+1)^2}$ and how to calculate it?
 A: Expand the factors in the numerator and in the denominator, then you'll have:
$$\lim_{n\to \infty}  \frac{(n^2-9)(3n+1)^2}{(n+3)(n+1)^3} = \lim_{n\to\infty} \frac{9n^4 + \ldots}{n^4 + \ldots} = 9,$$
where I have divided each term by $n^4$.
Cheers.
A: Write
$$\begin{array}{rcl}
\lim_{n \to \infty}\frac{(n^2-9)/(n+1)^3}{(n+3)/(3n+1)^2} &=& \lim_{n \to \infty}\frac{(n-3)\cdot(n+3)\cdot(3n+1)^2}{(n+3)\cdot(n+1)^3}\\
 &=& \lim_{n \to \infty}\frac{(n-3)\cdot(3n+1)^2}{(n+1)^3}\\
 &=& \lim_{n \to \infty}\frac{9n^3 -21n^2-17n-3}{n^3+3n^2+3n+1} \\
&=& \lim_{n \to \infty}\frac{9 -21n^{-1}-17n^{-2}-3n^{-3}}{1+3n^{-1}+3n^{-2}+1n^{-3}} \\
\end{array}$$
Thus the limit is $9$

Alternatively you can split up the fraction
$$\begin{array}{rcl}
\lim_{n \to \infty}\frac{(n^2-9)/(n+1)^3}{(n+3)/(3n+1)^2} &=& \lim_{n \to 
\infty}\frac{(n-3)\cdot(n+3)\cdot(3n+1)^2}{(n+3)\cdot(n+1)^3}\\
 &=& \lim_{n \to \infty}\frac{(n-3)\cdot(3n+1)^2}{(n+1)^3}\\
 &=& \lim_{n \to \infty} 9 - \frac{48}{n+1}+\frac{52}{(n+1)^2}-\frac{16}{(n+1)^3}\\
 &=& 9
\end{array}$$
