with proof find the $\displaystyle lim_{x -> 0+}(x^{x})$

my guess would be 1 as it is between $\displaystyle 0^0< x^x < 1^x for 0 < x < 1 $ and by the sandwich rule it convergrs to 1

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- May 26th 2009, 08:02 AM #1

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- May 26th 2009, 10:05 AM #2

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- May 26th 2009, 10:21 AM #3

- May 26th 2009, 02:43 PM #4

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The limit is of the indeterminate form $\displaystyle 0^0$

So we will 1st take the log of $\displaystyle x^x$ and then apply L'Hospital's rule.

AND

$\displaystyle \lim_{x\rightarrow 0^+}{\ln x^x} = \lim_{x\rightarrow 0^+}({x\ln x}) = \lim_{x\rightarrow 0^+}{\frac{\ln x}{\frac{1}{x}}} = \lim_{x\rightarrow 0^+}{\frac{\frac{1}{x}}{\frac{-1}{x^2}}} = \lim_{x\rightarrow 0^+}({-x}) = 0$.

Since $\displaystyle \ln x^x$ tends to 0, $\displaystyle x^x$ must tend to 1.

- May 26th 2009, 03:51 PM #5

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- May 26th 2009, 05:55 PM #6

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- May 26th 2009, 08:06 PM #7

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- May 26th 2009, 08:43 PM #8

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- Jun 13th 2009, 01:19 PM #9

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Hey, just another slightly different approach without using L'Hopital:

By considering their graphs it's easy to see that

$\displaystyle

\lim_{a \to - \infty} \int_{a}^0 \ e^x dx = - \lim_{b \to 0} \int_{b}^1 \ \ln{x} dx

$

Then the left hand side is easy to calculate:

$\displaystyle

\lim_{a \to - \infty} \int_{a}^0 \ e^x dx = \lim_{a \to - \infty} [e^x]_{a}^0 = 1

$

So on the right hand side

$\displaystyle

\lim_{b \to 0} \int_{b}^1 \ \ln{x} dx = -1 $

$\displaystyle

\lim_{b \to 0} [x \ln{x} - x]_{b}^1 = -1

$

rearraging we arrive at

$\displaystyle

\lim_{x \to 0} x \ln{x} = 0

\Rightarrow \lim_{x \to 0} x^x = 1

$

Using L'Hopital is clearly a lot quicker

Hope this was useful.

- Jun 13th 2009, 02:45 PM #10

- Jun 14th 2009, 12:28 AM #11

- Jun 14th 2009, 01:01 AM #12

- Jun 14th 2009, 11:45 PM #13
Some years ago I demonstrated in rigorous way that the expression $\displaystyle 0^{0}$ is not at all 'undetermined' but it is...

$\displaystyle 0^{0}=1$

That meas that ...

$\displaystyle f(x)= x^{x}$

... is continous in $\displaystyle x=0$ because the limit and the value of the function are identical...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

- Jun 14th 2009, 11:52 PM #14
I like the way you refer to yourself... It would be more interesting to give an outline of the proof... (I've seen some, I'm thinking about anyone who reads this)

That meas that ...

$\displaystyle f(x)= x^{x}$

... is continous in $\displaystyle x=0$ because the limit and the value of the function are identical...

- Jun 15th 2009, 01:30 AM #15
As Dante Alighieri wrote '... perder tempo a chi pił sa pił spiace...', so that the 'proof' I'll give to you is quite elementary...

phase 1: you activate the function 'calculator' in the 'accessories' of Windows...

phase 2: you set the calculator as 'scientific' ...

phase 3 : push in order '0' -> 'x^y' -> '0' -> '='...

After phase 3 in the display of calculator you will read $\displaystyle 1$, so that...

$\displaystyle 0^{0}=1$

Please avoid tedious further discussions ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$