May 2nd, 2016, 06:49 PM  #1 
Newbie Joined: Mar 2016 From: georgia Posts: 2 Thanks: 0  Indeterminate Form of [0^0]
Can anyone tell me if there is a function where the immediate indeterminate form is of [0^0] but after L'Hopital's rule, it reduces to a number. However, it can't reduce to the value of 1. Any other value will work. Thanks.

May 2nd, 2016, 08:02 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra 
$$\lim_{x \to 0} x^{k \over \log x} = \mathrm e^k$$
Last edited by v8archie; May 2nd, 2016 at 08:20 PM. 
May 2nd, 2016, 08:09 PM  #3 
Member Joined: Apr 2015 From: USA Posts: 46 Thanks: 32 
v8archie, maybe you meant to write $\displaystyle\lim_{x\to0}x^{\frac{\log k}{\log x}}=k$ 
May 2nd, 2016, 08:20 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra 
Not quite, but thanks for pointing out the error.

May 3rd, 2016, 12:06 AM  #5 
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271 
If k is a constant then ln(k) and exp(k) are also constants so there is no point having a fancy expression of the constant.

May 3rd, 2016, 01:37 AM  #6 
Newbie Joined: May 2016 From: italy Posts: 22 Thanks: 0 
the tricky part is this if you have some sort of f(x)^g(x) that goes to 0^0 or 1^0 or infinity^infinity in the most cases you can do e^g(x)lim f(X) and then do your derivite in some cases after this step you going to have another form of indeterminant like 0 over 0 or infinity over infinity and then you can apply hopital rule best wishes 
May 3rd, 2016, 11:40 AM  #7 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  $\displaystyle On \ \ the \ \ other \ \ hand, \ \ 1^{\infty} \ \ and \ \ {\infty}^0 \ \ $ are indeterminate forms.

May 3rd, 2016, 01:45 PM  #8 
Newbie Joined: May 2016 From: italy Posts: 22 Thanks: 0 
yes exactly sorry about that and thanks to mentioned


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