$$\lim_{x\to\infty} \frac{x}{\sqrt{9x^2 +x} -3}$$
I don't know what to do after this point. Also, the limit laws do not make sense to me when it comes to radicals or fractions. Do I evaluate now and treat the root as a fraction, thus it will be $0$? And what do I do with the $+3x$. Does it go in front of the limit ($3 \lim x$) and is being multiplied by the $3$ times infinity?
What is really happening?
Best Answer
$$\lim_{x\to\infty} \frac{x}{\sqrt{x^2(9 +\frac{1}{x})} -x\frac{3}{x}} =\lim_{x\to\infty} \frac{x}{x\sqrt{(9 +\frac{1}{x})} -x\frac{3}{x}}=$$ $$=\lim_{x\to\infty} \frac{1}{\sqrt{(9 +\frac{1}{x})} -\frac{3}{x}} = \frac{1}{\sqrt{9}} = \frac{1}{3}$$ since both $\frac{1}{x}$ and $\frac{3}{x}$ goes to $0$ as $x$ goes to $\infty$.