# The limit (and function) changes after rationalizing?

I want to evaluate the following: $$\lim_{r \rightarrow 0}\frac{-r^2}{2 \left(\sqrt{1-\frac{r^2}{4}}-1 \right)}$$

I look at the graph and see that it seems to be going to zero. This makes sense to me because if I replace r with zero this function is defined and continuous near zero and the value of the function is zero.

So, I think: $$\lim_{r \rightarrow 0}\frac{-r^2}{2 \left(\sqrt{1-\frac{r^2}{4}}-1 \right)}=0$$

But next I try something else I rationalize the denominator since this is a good technique for solving limits. $$\lim_{r \rightarrow 0}\frac{-r^2 \left(\sqrt{1-\frac{r^2}{4}}+1 \right)}{2 \left(1-\frac{r^2}{4}-1 \right)}$$

Then, $$\lim_{r \rightarrow 0}\frac{-r^2 \left(\sqrt{1-\frac{r^2}{4}}+1 \right)}{\frac{-r^2}{2}}$$ so... $$\lim_{r \rightarrow 0} 2 \left(\sqrt{1-\frac{r^2}{2}}+1 \right)=4$$

What have I done? How can rationalizing change the graph? My guess is that the denominator $2 \left(\sqrt{1-\frac{r^2}{2}}-1 \right)$ is "divisible by $r^2$ in some not obvious way?

I know my original reasoning was sloppy, (not a proof) but the fact that the graph shows a limit of zero has me very confused.

How do I avoid this error? Just always rationalize everything? Why would I do that?

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I gave a +1 for "Why would I do that?" :) – Srivatsan Sep 2 '11 at 4:07

Beyond that, when $r=0$, the numerator of the expression is $-r^2=-0^2=0$ and the denominator is $2 \left(\sqrt{1-\frac{r^2}{4}}-1 \right)=2 \left(\sqrt{1-\frac{0^2}{4}}-1 \right)=2 \left(\sqrt{1}-1 \right)=0$—that is, the expression is of the form $\frac{0}{0}$, which is an indeterminate form, so it needs further investigation.