# What are the indeterminate forms such that L'Hopital's rule can be used?

• Jan 31st 2011, 08:29 PM
crossbone
What are the indeterminate forms such that L'Hopital's rule can be used?
$\displaystyle \frac{0}{0}, \frac{\infty}{\infty}, 0\times{\infty},{\infty}^{\infty}, {\infty}^0,0^0, 1^{\infty}, \infty - \infty.$

How about $\displaystyle 0^{\infty}$? anything else?
• Jan 31st 2011, 08:33 PM
dwsmith
Quote:

Originally Posted by crossbone
$\displaystyle \frac{0}{0}, \frac{\infty}{\infty}, 0\times{\infty},{\infty}^{\infty}, {\infty}^0,0^0, 1^{\infty}, \infty - \infty.$

How about $\displaystyle 0^{\infty}$? anything else?

Check wikipedia or a textbook.
• Jan 31st 2011, 09:09 PM
Prove It
You can go directly to L'Hospital's Rule if it's of the form $\displaystyle \displaystyle \frac{0}{0}$ or $\displaystyle \displaystyle \frac{\infty}{\infty}$.

However, you can apply some transformations to the other forms to get them to $\displaystyle \displaystyle \frac{0}{0}$ or $\displaystyle \displaystyle \frac{\infty}{\infty}$.
• Jan 31st 2011, 11:28 PM
chisigma
Quote:

Originally Posted by crossbone
$\displaystyle \frac{0}{0}, \frac{\infty}{\infty}, 0\times{\infty},{\infty}^{\infty}, {\infty}^0,0^0, 1^{\infty}, \infty - \infty.$

How about $\displaystyle 0^{\infty}$? anything else?

The form $\displaystyle 0^{+ \infty}$ is not 'indeterminate'!... more precisely if You have a function like $\displaystyle y(x)= f(x)^{g(x)}$ with $\displaystyle \lim_{x \rightarrow x_{0}} f(x)=0$ and $\displaystyle \lim_{x \rightarrow x_{0}} g(x)= +\infty$ is $\displaystyle \lim_{x \rightarrow x_{0}} f(x)^{g(x)}=0$...

Are You sure that $\displaystyle \infty^{\infty}$ is 'indeterminate'?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Feb 1st 2011, 01:29 AM
chisigma
Other 'indeterminate forms' that are not reported in the 'holy books' and can be solved with l'Hopital's rule are [in my opinion...] $\displaystyle \log_{0} 0$ and $\displaystyle \log_{\infty} \infty$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Feb 1st 2011, 03:40 AM
crossbone
Quote:

Originally Posted by dwsmith
Check wikipedia or a textbook.

I did but it didn't include $\displaystyle 0^{\infty}$ even though $\displaystyle {\infty}^0$ and $\displaystyle 0^0$ were. isn't it an indeterminate form?
• Feb 1st 2011, 04:10 AM
skeeter
First paragraph from this source ...

http://spectrum.troy.edu/~andrew/doc...ETERMINATE.pdf

... lists "seven known" indeterminate forms.
• Feb 1st 2011, 06:31 AM
crossbone
seems like it's not then. so what's the value of $\displaystyle 0^{\infty} ? 0?$
• Feb 1st 2011, 07:01 AM
skeeter
Quote:

Originally Posted by crossbone
seems like it's not then. so what's the value of $\displaystyle 0^{\infty} ? 0?$

consider ...

$\displaystyle \displaystyle \lim_{n \to \infty} r^n$ when $\displaystyle |r| < 1$
• Feb 1st 2011, 07:18 AM
crossbone
makes sense. thanks!