Find the value of : $\lim_{x \to \infty}( \sqrt{4x^2+5x} - \sqrt{4x^2+x})$

As $x\to\infty$ $$\displaystyle \frac{4x}{\sqrt{4x^2+5x} + \sqrt{4x^2+x}}=\displaystyle \frac{4}{\sqrt{4+\dfrac{5}{x}} + \sqrt{4+\dfrac{1}{x}}}\to1$$

Set $h=\dfrac1x$

$$4x^2+ax=\frac{4+ah}{h^2}\implies\sqrt{4x^2+ax}=\frac{\sqrt{4+ah}}{\sqrt{h^2}}$$

Now as $h\to0^+,h>0\implies\sqrt{h^2}=|h|=h$

$$\implies\lim_{x \to \infty} (\sqrt{4x^2+5x} - \sqrt{4x^2+x})=\lim_{h\to0^+}\frac{\sqrt{4+5h}-\sqrt{4+h}}h$$

$$=\lim_{h\to0^+}\frac{4+5h-(4+h)}{h(\sqrt{4+5h}+\sqrt{4+h})}=\cdots$$

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 $\frac{5}{4}-\frac 14=1$.