A graph won't help much with this limit!

$\begin{aligned} L & = \lim_{x \to \infty} \frac {\log_{27} x^2 + 5}{\log_9 x} & \small \color{#3D99F6} \text{Change all log to base 3} \\ & = \lim_{x \to \infty} \frac {\frac {\log_3 x^2}{\log_3 27} + 5}{\frac {\log_3 x}{\log_3 9}} \\ & = \lim_{x \to \infty} \frac {\frac {2 \log_3 x}3 + 5}{\frac {\log_3 x}2} \\ & = \lim_{x \to \infty} \frac {4\log_3 x + 30}{3\log_3 x} & \small \color{#3D99F6} \text{Divide up and down by }\log_3 x \\ & = \lim_{x \to \infty} \frac {4 + \frac {30}{\log_3 x}}3 \\ & = \frac 43 \end{aligned}$

Therefore, $A+B = 4+3 = \boxed 7$ .

First, using the change of base formula, we may write:

$\log_{27}(x^2) = \log_9(x^2) / \log_9 (27) \\ = \log_9(x^2) / (3/2) \\ = 2/3 \log_9(x^2)$

We may thus rewrite our expression as

$\frac{2/3 \log_9(x^2) + 5}{\log_9(x)}$

Next, using log properties, we may rewrite $\log_9(x^2) = 2 \log_9(x)$ , and we get our expression is equivalent to:

$\frac{4/3 \log_9(x) + 5}{\log_9(x)}$

Multiplying through by $1/\log_9(x)$ in the numerator and the denominator we have:

$\frac{4/3 + 5/\log_9(x)}{1}$

As $x \rightarrow \infty$ , we know $\log_9(x) \rightarrow \infty$ , thus $1/\log_9(x) \rightarrow 0$ . Therefore the limit we want is $4/3$ and our answer is 4+3=7.