Limit of a function with square roots

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?

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.

• Can you explain how you go from $\frac {t(1-t^3)}{1-t}$ to $t(1+t+t^2)$? May 8, 2014 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. May 8, 2014 at 3:03
• woah, that makes sense. Thanks a lot! May 8, 2014 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. May 8, 2014 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 $s\to 1$.

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

• I haven't learned about derivatives yet, so this doesn't really help. Thanks though! May 8, 2014 at 2:50
• @Joseph But if you haven't learned about limits, then how is this a question about limits?
– user122283
May 8, 2014 at 2:51
• I meant derivatives, sorry. Corrected my reply. May 8, 2014 at 2:57

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

• $\sqrt{s} = s^3$????? May 8, 2014 at 4:53