Find the value of : $\lim_{x \to \infty}( \sqrt{4x^2+5x} - \sqrt{4x^2+x})$ $$\lim_{x \to \infty} \left(\sqrt{4x^2+5x} - \sqrt{4x^2+x}\ \right)$$
I have a lot of approaches, but it seems that I get stuck in all of those unfortunately. So for example I have tried to multiply both numerator and denominator by the conjugate $\left(\sqrt{4x^2+5x} + \sqrt{4x^2+x}\right)$, then I get $\displaystyle \frac{4x}{\sqrt{4x^2+5x} + \sqrt{4x^2+x}}$, but I can conclude nothing out of it. 

Since you already received answers, let me show you another approach you could use. $$A=\sqrt{4x^2+5x} - \sqrt{4x^2+x}=2x \sqrt{1+\frac{5}{4x}}-2x \sqrt{1+\frac{1}{4x}}=2x \Big(\sqrt{1+\frac{5}{4x}}-\sqrt{1+\frac{1}{4x}}\Big)$$ Now, you may be already know that, when $y$ is small compared to $1$ $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ Use this twice, replacing $y$ by $\frac{5}{4x}$ for the first radical and by $\frac{1}{4x}$ for the second radical. You will then have $$A=2x \Big(\frac{1}{2 x}-\frac{3}{16 x^2}+\cdots\Big)=1-\frac{3}{8 x}+\cdots$$ which shows the limit and how it is approached.
