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. 

Note that $$\sqrt{4x^2+5x}-\sqrt{4x^2+x} = \frac{4x}{\sqrt{4x^2+5x}+\sqrt{4x^2+x}}$$
And the denominator is between $2x+2x=4x$ and $(2x+\frac{5}{4})+(2x+\frac{1}{4})=4x+\frac{3}{2}$. So the limit be $1$.
Alternatively, show that:
$$\lim \left(2x+\frac{5}{4}-\sqrt{4x^2+5x}\right) = 0$$
and
$$\lim \left(2x+\frac{1}{4}-\sqrt{4x^2+x}\right) = 0$$
Then again deduce that  the limit of the difference must be $0$, so the limit you are seeking is $\frac{5}{4}-\frac 14=1$.
