$$\lim_{x \to \infty} \left(\sqrt{4x^2+5x} - \sqrt{4x^2+x}\ \right)$$

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.

  • 2
    $\begingroup$ there is a mistake in the denominator, you should have a '+' $\endgroup$ – Mosk Dec 27 '14 at 13:19
  • 1
    $\begingroup$ Divide the numerator and the denominator by $x$. $\endgroup$ – user35603 Dec 27 '14 at 13:21
  • $\begingroup$ My suitcase of dirty tricks says $\sqrt{ (2x+\tfrac54)^2 + \epsilon} - \sqrt{ (2x+\tfrac14)^2 + \epsilon'} = \tfrac54-\tfrac14$. $\endgroup$ – Myself Dec 27 '14 at 13:22
  • $\begingroup$ @DheerajKumar, thanks :) $\endgroup$ – user35603 Dec 27 '14 at 13:23

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

  • $\begingroup$ +1 How could I miss this? So simple.. $\endgroup$ – Gary Dec 27 '14 at 13:28

Set $h=\dfrac1x$


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

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



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:

$$\lim \left(2x+\frac{5}{4}-\sqrt{4x^2+5x}\right) = 0$$


$$\lim \left(2x+\frac{1}{4}-\sqrt{4x^2+x}\right) = 0$$

Then again deduce that the limit of the difference must be $0$, so the limit you are seeking is $\frac{5}{4}-\frac 14=1$.


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.


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