# How to solve $\lim _{x\to \infty }\left(x\left(\sqrt{9x^2+5}-3x\right)\right)$?

I tried with Wolfram Alpha, but it fails to provide a step-by-step solution and gives me only the answer.

• Taylor series if you know them, rationalization (multiply by $\frac{\sqrt{9x^2+5}+3x}{\sqrt{9x^2+5}+3x}$) if you don't? – Clement C. Nov 21 '16 at 15:09

Rewrite $\sqrt{9x^2+5}-3x$ as a fraction.
Alternative hint: set $x=1/\sqrt{t}$ and compute $$\lim_{t\to0^+}\frac{1}{\sqrt{t}}\left(\sqrt{\frac{9}{t}+5}-\frac{3}{\sqrt{t}}\right) = \lim_{t\to0^+}\frac{\sqrt{9+5t}-3}{t}$$ which is a simple derivative.
$\lim_{x\to \infty }\left(x\left(\sqrt{9x^2+5}-3x\right)\right)=\lim_{x\to\infty}\left(x\left(\sqrt{9x^2+5}-3x\right)\frac{\sqrt{9x^2+5}+3x}{\sqrt{9x^2+5}+3x}\right)=\lim_{x\to\infty}\frac{5x}{\sqrt{9x^2+5}+3x}=5/6$