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?
 A: 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 $s\to 1$.
A: 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.
A: $$\lim_{s\to 1}\dfrac{\sqrt{s}-s^2}{1-\sqrt{s}}=\lim_{s\to 1}\dfrac{\dfrac{1}{2\sqrt{s}}-2s}{-\dfrac{1}{2\sqrt{s}}}=\lim_{s\to 1}\dfrac{\dfrac{1-4s\sqrt{s}}{2\sqrt{s}}}{-\dfrac{1}{2\sqrt{s}}}=\lim_{s\to 1}(-1+4s\sqrt{s})$$
By L'Hopital's rule.
A: $$\frac{\sqrt{s} - s^2}{1 - \sqrt{s}} = \frac{\sqrt s \left(1 - s^{3/2}\right)}{\left(1 - s^{3/2}\right)\left(1 + s^{3/2}\right)} = \frac{\sqrt s}{1 + s^{3/2}}$$
