Finding the limit

Quote:

Originally Posted by johnsy123

I am not sure how to prove this by writing it out. I graphed it and the limit of that is 1/3.
$\displaystyle \lim_{ x \to \infty }{ \frac{ x - 3 }{ \sqrt{ 9x^2 - 1 } } }$

$\displaystyle \begin{align*} \lim_{x \to \infty}\frac{x - 3}{\sqrt{9x^2 - 1}} &= \lim_{x \to \infty}\frac{\frac{1}{x}\left(x - 3\right)}{\frac{1}{x}\left(\sqrt{9x^2 - 1}\right)} \\ &= \lim_{x \to \infty}\frac{1 - \frac{3}{x}}{\sqrt{\frac{1}{x^2}}\,\sqrt{9x^2 - 1}} \\ &= \lim_{x \to \infty}\frac{1 - \frac{3}{x}}{\sqrt{\frac{9x^2 - 1}{x^2}}} \\ &= \lim_{x \to \infty}\frac{1 - \frac{3}{x}}{\sqrt{9 - \frac{1}{x^2}}} \\ &= \frac{1 - 0}{\sqrt{9 - 0}} \\ &= \frac{1}{\sqrt{9}} \\ &= \frac{1}{3} \end{align*}$