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

• Jan 31st 2011, 09:29 PM
crossbone
What are the indeterminate forms such that L'Hopital's rule can be used?
$\frac{0}{0}, \frac{\infty}{\infty}, 0\times{\infty},{\infty}^{\infty}, {\infty}^0,0^0, 1^{\infty}, \infty - \infty.$

How about $0^{\infty}$? anything else?
• Jan 31st 2011, 09:33 PM
dwsmith
Quote:

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

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

Check wikipedia or a textbook.
• Jan 31st 2011, 10:09 PM
Prove It
You can go directly to L'Hospital's Rule if it's of the form $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$.

However, you can apply some transformations to the other forms to get them to $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$.
• Feb 1st 2011, 12:28 AM
chisigma
Quote:

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

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

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

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

Kind regards

$\chi$ $\sigma$
• Feb 1st 2011, 02:29 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...] $\log_{0} 0$ and $\log_{\infty} \infty$...

Kind regards

$\chi$ $\sigma$
• Feb 1st 2011, 04:40 AM
crossbone
Quote:

Originally Posted by dwsmith
Check wikipedia or a textbook.

I did but it didn't include $0^{\infty}$ even though ${\infty}^0$ and $0^0$ were. isn't it an indeterminate form?
• Feb 1st 2011, 05:10 AM
skeeter
First paragraph from this source ...

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

... lists "seven known" indeterminate forms.
• Feb 1st 2011, 07:31 AM
crossbone
seems like it's not then. so what's the value of $0^{\infty} ? 0?$
• Feb 1st 2011, 08:01 AM
skeeter
Quote:

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

consider ...

$\displaystyle \lim_{n \to \infty} r^n$ when $|r| < 1$
• Feb 1st 2011, 08:18 AM
crossbone
makes sense. thanks!