Solve $\lim_{x \rightarrow +\infty}[\sqrt{2x^2+x-1}-\sqrt{2(x^2+x)} ]$ without using L'Hopital

Question

I tried:

$$\lim_{x \rightarrow +\infty}[\sqrt{2x^2+x-1}-\sqrt{2(x^2+x)} ] = \\ \sqrt{(2x-1)(x+1)}-\sqrt{2x(x+1)}= \\ \sqrt{2x-1}\cdot \sqrt{x+1}-\sqrt{2x}\cdot \sqrt{x+1}=\\ \sqrt{(x+1)}(\sqrt{(2x-1)}-\sqrt{(2x)})= \\ ???$$

What do I do next? How do I solve this?

Answer 1

One may write, as $x \to \infty$, $$ \begin{align} &\sqrt{2x^2+x-1}-\sqrt{2(x^2+x)} \\\\&=\frac{[\sqrt{2x^2+x-1}-\sqrt{2(x^2+x)}][\sqrt{2x^2+x-1}+\sqrt{2(x^2+x)}] }{\sqrt{2x^2+x-1}+\sqrt{2(x^2+x)}} \\\\&=-\frac{x+1}{\sqrt{2x^2+x-1}+\sqrt{2(x^2+x)}} \\\\&=-\frac{1+1/x}{\sqrt{2+1/x-1/x^2}+\sqrt{2+2/x}} \end{align} $$ then one may conclude easily.