Computing $\lim_{x\to\infty}{3^x\over \sqrt {9^x - 4^x}}$ 
Compute$$\lim_{x\to\infty}{3^x\over \sqrt {9^x - 4^x}}$$

Can one use end behavior to solve this? I.E. 
$$\lim_{x\to\infty}{\sqrt{3^{2x}\over (3^{2x} - 2^{2x})}}$$
and therefore divide $3^{2x}$ by $3^{2x}$?  leaving us with an anwser of $1$? 

For his problem exist simple solution.
$$ \lim_{x \to \infty^{+}} \frac{3^x}{\sqrt{9^x - 4^x}} 
= \lim_{x \to \infty^{+}} \frac{3^x}{\sqrt{(3^2)^x( 1 - \frac{4^x}{9^x})}}
= \lim_{x \to \infty^{+}} \frac{3^x}{\sqrt{(3^x)^2} \cdot \sqrt{1 - \frac{4^x}{9^x}}} = \\
= \lim_{x \to \infty^{+}} \frac{3^x}{3^x \cdot \sqrt{1 - \left(\frac{4}{9}\right)^x}} = \lim_{x \to \infty^{+}} \frac{1}{\sqrt{1 - \left(\frac{4}{9}\right)^x}} = 1$$
I used here fact, that $|q| < 1 \Rightarrow \lim_{x \to \infty}q^x=0$. In fact it's same to yours, but I excluded $3^x$ from root, you just included.
