Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've got the following limit to solve:

$$\lim_{s\to 1} \frac{\sqrt{s}-s^2}{1-\sqrt{s}}$$

I was taught to multiply by the conjugate to get rid of roots, but that doesn't help, or at least I don't know what to do once I do it. I can't find a way to make the denominator not be zero when replacing $s$ for $1$. Help?

share|cite|improve this question
up vote 10 down vote accepted

Try putting $t=\sqrt s$ to get $$\frac {t-t^4}{1-t}=\frac {t(1-t^3)}{1-t}=t(1+t+t^2)$$You can notice this without the substitution, of course, but sometimes a substitution like this helps to clarify what is going on.

share|cite|improve this answer
Can you explain how you go from $\frac {t(1-t^3)}{1-t}$ to $t(1+t+t^2)$? – Juan José Castro May 8 '14 at 2:51
@Joseph $1-t^3=(1-t)(1+t+t^2)$ as you can check by direct multiplication. The factorisation $1-t^n=(1-t)(1+t+t^2+\dots +t^{n-1})$ is a standard one you should know. It is the same as summing a finite geometric progression $$1+t+t^2+ \dots +t^{n-1}=\frac {1-t^n}{1-t}$$. The polynomials $p_n(t)=t^n-1$ have roots the $n^{th}$ roots of unity and are really important in algebra and number theory. – Mark Bennet May 8 '14 at 3:03
woah, that makes sense. Thanks a lot! – Juan José Castro May 8 '14 at 3:04
@Joseph I should perhaps add that since the fraction goes to $\frac 00$ and is a rational function of $t=\sqrt s$ you should expect a factor of $t-1=\sqrt s -1$ or $1-t=1-\sqrt s$ to cancel from numerator and denominator. This happens because $t=1$ is a root of the numerator expression and the denominator expression. L'Hopital's rule works more generally - e.g. when you have exponentials, logarithms and trigonometric functions. – Mark Bennet May 8 '14 at 3:09

You were on the right track using the conjugate.

$$\begin{align*} \frac{\sqrt{s}-s^2}{1-\sqrt{s}}&=\frac{\sqrt{s}-s^2}{1-\sqrt{s}}\times\frac{1+\sqrt{s}}{1+\sqrt{s}}\\ &=\frac{\sqrt{s}-s^2+s-s^{5/2}}{1-s}\\ &=\frac{s-s^2+\sqrt{s}-s^{5/2}}{1-s}\\ &=\frac{s(1-s)+\sqrt{s}(1-s^{2})}{1-s}\\ &=\frac{s(1-s)+\sqrt{s}(1-s)(1+s)}{1-s}\\ &=s+\sqrt{s}(1+s)\\ \end{align*}$$

This is now ready for taking the limit as $s\to 1$.

share|cite|improve this answer

$$\lim_{s\to 1}\dfrac{\sqrt{s}-s^2}{1-\sqrt{s}}=\lim_{s\to 1}\dfrac{\dfrac{1}{2\sqrt{s}}-2s}{-\dfrac{1}{2\sqrt{s}}}=\lim_{s\to 1}\dfrac{\dfrac{1-4s\sqrt{s}}{2\sqrt{s}}}{-\dfrac{1}{2\sqrt{s}}}=\lim_{s\to 1}(-1+4s\sqrt{s})$$ By L'Hopital's rule.

share|cite|improve this answer
I haven't learned about derivatives yet, so this doesn't really help. Thanks though! – Juan José Castro May 8 '14 at 2:50
@Joseph But if you haven't learned about limits, then how is this a question about limits? – Sanath K. Devalapurkar May 8 '14 at 2:51
I meant derivatives, sorry. Corrected my reply. – Juan José Castro May 8 '14 at 2:57

$$\frac{\sqrt{s} - s^2}{1 - \sqrt{s}} = \frac{\sqrt s \left(1 - s^{3/2}\right)}{\left(1 - s^{3/2}\right)\left(1 + s^{3/2}\right)} = \frac{\sqrt s}{1 + s^{3/2}}$$

share|cite|improve this answer
$\sqrt{s} = s^3$????? – 6005 May 8 '14 at 4:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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