[College Calc 01-02. Limits of Sequences] #01

Notice that the denominator is $9n^2+27n+27.$ Dividing the numerator and denominator by $n^2$ we have

$\cdots = \lim_{n\to\infty}\frac{\sqrt{25+\dfrac{16}{n^2}+\dfrac{9}{n^4}}}{9+\dfrac{27}{n}^{\phantom{2}}\!\!\!+\dfrac{27}{n^2}} = \frac{5}{9}.$

Hence $p+q=5+9=\boxed{14}.$

$\begin{aligned} L & = \lim_{n \to \infty} \frac {\sqrt{25n^4+16n^2 + 9}}{(n+3)^3-n^3} \\ & = \lim_{n \to \infty} \frac {\sqrt{25n^4+16n^2 + 9}}{n^3+9n^2 + 27n + 27 -n^3} \\ & = \lim_{n \to \infty} \frac {\sqrt{25n^4+16n^2 + 9}}{9n^2 + 27n + 27} & \small \blue{\text{Divide up and down by }n^2} \\ & = \lim_{n \to \infty} \frac {\sqrt{25+ \frac {16}{n^2} + \frac 9{n^4}}}{9 + \frac {27}n + \frac {27}{n^2}} \\ & = \frac 59 \end{aligned}$

Therefore $p+q = 5 + 9 = \boxed{14}$ .