Limit Problem with a Square Root

[Jason76]

I pretty much got "limits" down pat, but a few snags here and there:

\(\displaystyle \lim\) as \(\displaystyle r\) approaches \(\displaystyle 1\) is \(\displaystyle \dfrac{1 - r^{3}}{2 - \sqrt{r^{2} + 3}}\) - Plugging in we get an indeterminate value \(\displaystyle \dfrac{0}{0}\)so we have to try either factoring and then "L' Hospital's rule" or just L' Hospitals rule (This rule involves differentiating and then substituting the value under the limit sign). But the problem lies with the square root. We cannot do the rule on it, but we can on everything else.


\(\displaystyle \lim\) as \(\displaystyle r\) approaches \(\displaystyle 1\) is \(\displaystyle \dfrac{1 - r^{3}}{2 - \sqrt{r^{2} + 3}}\)

\(\displaystyle \lim\) as \(\displaystyle r\) approaches \(\displaystyle 1\) is \(\displaystyle \dfrac{-3r^{2}}{?}\) - We differentiated on the top. What goes in the bottom and why? Some hints?

 

[DrPhil]

I pretty much got "limits" down pat, but a few snags here and there:

\(\displaystyle \lim\) as \(\displaystyle r\) approaches \(\displaystyle 1\) is \(\displaystyle \dfrac{1 - r^{3}}{2 - \sqrt{r^{2} + 3}}\) - Plugging in we get an indeterminate value \(\displaystyle \dfrac{0}{0}\)so we have to try either factoring and then "L' Hospital's rule" or just L' Hospitals rule (This rule involves differentiating and then substituting the value under the limit sign). But the problem lies with the square root. We cannot do the rule on it, but we can on everything else.


\(\displaystyle \lim\) as \(\displaystyle r\) approaches \(\displaystyle 1\) is \(\displaystyle \dfrac{1 - r^{3}}{2 - \sqrt{r^{2} + 3}}\)

\(\displaystyle \lim\) as \(\displaystyle r\) approaches \(\displaystyle 1\) is \(\displaystyle \dfrac{-3r^{2}}{?}\) - We differentiated on the top. What goes in the bottom and why? Some hints?

Rationalize the denominator by multiplying by
\(\displaystyle \dfrac{2 + \sqrt{r^{2} + 3}}{2 + \sqrt{r^{2} + 3}}\)
You will then find there is a common factor of \(\displaystyle (1 - r)\) in both the numerator and denominator. Canceling that factor may be all you need to do to have an expression you can evaluate at \(\displaystyle r = 1\).

 

[Jason76]

Rationalize the denominator by multiplying by
\(\displaystyle \dfrac{2 + \sqrt{r^{2} + 3}}{2 + \sqrt{r^{2} + 3}}\)
You will then find there is a common factor of \(\displaystyle (1 - r)\) in both the numerator and denominator. Canceling that factor may be all you need to do to have an expression you can evaluate at \(\displaystyle r = 1\).

That's one way to do it, but i wonder if substitution also works?

 

[DrPhil]

That's one way to do it, but i wonder if substitution also works?

What did you have in mind to substitute? That is not a standard method to evaluate a limit. Can you demonstrate what you mean? There may well be other methods to use - for instance following through on l'Hospital by differentiating the denominator. I don't know why you say "can't take the rule on it." The derivative may be messy, but when you take the limit r-->1 it will be non-zero.

What IS a standard method is to rationalize to eliminate the square root in the denominator. Please follow our suggestions - I am getting to the point of not bothering to answer your questions because you are so reluctant to follow our advice. If you want to argue one method is easier than another, do it both ways to show us.