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May 2nd, 2016, 06:49 PM   #1
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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.
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May 2nd, 2016, 08:02 PM   #2
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$$\lim_{x \to 0} x^{k \over \log x} = \mathrm e^k$$

Last edited by v8archie; May 2nd, 2016 at 08:20 PM.
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May 2nd, 2016, 08:09 PM   #3
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v8archie, maybe you meant to write

$\displaystyle\lim_{x\to0}x^{\frac{\log k}{\log x}}=k$
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May 2nd, 2016, 08:20 PM   #4
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Not quite, but thanks for pointing out the error.
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May 3rd, 2016, 12:06 AM   #5
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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.
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May 3rd, 2016, 01:37 AM   #6
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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
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May 3rd, 2016, 11:40 AM   #7
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Quote:
Originally Posted by programmer View Post
the tricky part is this if you have some sort of f(x)^g(x) that goes to 0^0

or 1^0 $\displaystyle \ \ \ \ \ $That isn't an indeterminate form.

or infinity^infinity $\displaystyle \ \ \ \ \ $That isn't an indeterminate form.

$\displaystyle On \ \ the \ \ other \ \ hand, \ \ 1^{\infty} \ \ and \ \ {\infty}^0 \ \ $ are indeterminate forms.
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May 3rd, 2016, 01:45 PM   #8
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yes exactly sorry about that and thanks to mentioned
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