2
$\begingroup$

Evaluate: $\displaystyle\lim_{x \to \infty} (\sqrt{x+2}-\sqrt{x})$

I was given this problem. I'm not sure how to tackle it. $\infty-\infty$ is in indeterminate form, so I need to get it into a fraction form in order to solve. I did this by pulling an $x^2$ out of it: $$\lim_{x\to\infty}x\left(\sqrt{\frac1x+\frac2{x^2}}-\sqrt{\frac1x}\right)\\=\lim_{x\to\infty}\frac{\left(\sqrt{\frac1x+\frac2{x^2}}-\sqrt{\frac1x}\right)}{\frac1x}$$ Now taking the derivative: $$\lim_{x\to\infty}\Large\frac{\frac{\frac{-1}{x^2}+\frac{-4}{x^3}}{2\sqrt{\frac1x+\frac2{x^2}}}-\frac{-\frac1{x^2}}{2\sqrt{\frac1x}}}{-\frac1{x^2}}\\=\lim_{x\to\infty}\frac{-x^2\left(\frac{-1}{x^2}+\frac{1}{x^2}\right)}{2\sqrt{\frac1x+\frac2{x^2}}-2\sqrt{\frac1x}}\\=\lim_{x\to\infty}\frac{-x^2\left(0\right)}{2\sqrt{\frac1x+\frac2{x^2}}-2\sqrt{\frac1x}}\\=\lim_{x\to\infty}0=0$$ Since my numerator becomes zero, my entire fraction becomes zero, making zero the answer.

Is this correct? This was very messy - did I miss the easy way to do it? Is there a better way to do problems of this sort?

| cite | improve this question | | | | |
$\endgroup$
  • 1
    $\begingroup$ Your computation is NOT OK if the numerator becomes $0$ and the denominator also becomes $0$. $\endgroup$ – GEdgar Dec 30 '19 at 21:53
  • $\begingroup$ @GEdgar - here, my denominator also becomes zero. Does that mean my method of computation did not work? $\endgroup$ – Burt Dec 30 '19 at 21:54
  • $\begingroup$ You can have a look at: Prove $\lim_{x\to\infty} \left( \sqrt{x+1} - \sqrt{x} \right) = 0$. (Change from $x+1$ to $x+2$ does not make that much difference.) $\endgroup$ – Martin Sleziak May 15 at 13:11
9
$\begingroup$

Hint: $$\sqrt{x+2}-\sqrt{x}=\frac{2}{\sqrt{x+2}+\sqrt{x}}$$

| cite | improve this answer | | | | |
$\endgroup$
  • 2
    $\begingroup$ Wow, that was simple:) $\endgroup$ – Burt Dec 30 '19 at 21:48
  • 1
    $\begingroup$ Why is this true? $\endgroup$ – Burt Dec 30 '19 at 21:48
  • 2
    $\begingroup$ $(\sqrt{x+2}-\sqrt{x})(\sqrt{x+2}+\sqrt{x})=(x+2)-x=2$ $\endgroup$ – A. Goodier Dec 30 '19 at 21:49
  • 1
    $\begingroup$ This is the usual trick which is used when we have subtraction of square roots. It reminds of the well known trick of multiplying and dividing by a complex conjugate to get rid of the imaginary part. $\endgroup$ – Mark Dec 30 '19 at 21:55
  • 3
    $\begingroup$ Called "rationalize the numerator". $\endgroup$ – GEdgar Dec 30 '19 at 22:01
1
$\begingroup$

Your answer is correct. You could also use binomial series:

$$\lim_{x\to\infty}\left(\sqrt{x+2}-\sqrt x\right)=\lim_{x\to\infty}\sqrt x\left(\sqrt{1+\frac2x}-1\right)$$

$$=\lim_{x\to\infty}\sqrt{x}\left(1+\frac1x+o\left(\frac1x\right)-1\right)=\lim_{x\to\infty}\dfrac{\sqrt x}x=0.$$

| cite | improve this answer | | | | |
$\endgroup$
  • 1
    $\begingroup$ But would he get credit for a correct answer with an incorrect method? $\endgroup$ – GEdgar Dec 30 '19 at 21:54
0
$\begingroup$

We have that $\sqrt{x+2} - \sqrt{x} = \dfrac{(\sqrt{x+2}-\sqrt{x})(\sqrt{x+2}+\sqrt{x})}{\sqrt{x+2}+\sqrt{x}} = \dfrac{2}{\sqrt{x-2}+\sqrt{x}}.$ Hence $\lim\limits_{x\to\infty} \sqrt{x+2}-\sqrt{x} = \lim\limits_{x\to\infty}\dfrac{2}{\sqrt{x}(\sqrt{1-\frac{2}{x}}+1)}=0.$

If you were looking for a nonzero answer, you could have tried something like $\lim\limits_{x\to\infty} \sqrt{x^2+x}-x$, which evaluates to $\dfrac{1}{2}$.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ So was I totally wrong - my answer seems to really simplify to $\frac00$? $\endgroup$ – Burt Dec 30 '19 at 22:52
  • 5
    $\begingroup$ you got the wrong denominator for the third last line. $\endgroup$ – Simon Fraser Dec 30 '19 at 23:19
  • $\begingroup$ How did I mess up? $\endgroup$ – Burt Dec 31 '19 at 2:13
  • 5
    $\begingroup$ $\dfrac{\frac{-\frac{1}{x^2}-\frac{4}{x^3}}{2\sqrt{\frac{1}{x}+\frac{2}{x^2}}}-\frac{-\frac{1}{x^2}}{2\sqrt{\frac{1}{x}}}}{-\frac{1}{x^2}}\neq \dfrac{-x^2(-\frac{1}{x^2}+\frac{1}{x^2})}{2\sqrt{\frac{1}{x}+\frac{2}{x^2}}-2\sqrt{\frac{1}{x}}}$ $\endgroup$ – Simon Fraser Dec 31 '19 at 3:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.