1. ## lim x^x

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

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

2. Hi

Originally Posted by rosh3000
with proof find the $lim_{x -> 0+}(x^{x})$
I would solve it this way:

$lim \ x^x = lim \ e^{ln(x)*x} = e^0 = 1$

Originally Posted by rosh3000
my guess would be 1 as it is between $0^0< x^x < 1^x for 0 < x < 1$ and by the sandwich rule it convergrs to 1

Yours
Rapha

3. The sandwich theorem says :

$a_n\leq b_n\leq c_n$ and $\lim_{n\to\infty} a_n=\lim_{n\to\infty} c_n=l \quad \Rightarrow \lim_{n\to\infty} b_n=l$

What you used is false and not the sandwich theorem...

4. Originally Posted by rosh3000
with proof find the $lim_{x -> 0+}(x^{x})$

my guess would be 1 as it is between $0^0< x^x < 1^x for 0 < x < 1$ and by the sandwich rule it convergrs to 1
The limit is of the indeterminate form $0^0$

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

AND

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

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

5. thanks you mak it seem so easy

6. Originally Posted by Rapha
Hi

I would solve it this way:

$lim \ x^x = lim \ e^{ln(x)*x} = e^0 = 1$

Yours
Rapha

Does not the product (ln x)*x ,as x goes to zero go to $-\infty*0$,which is indeterminent,if im not mistaken??

7. $\lim_{x\to 0}x\ln(x)=0$. You can either do this by looking at the Taylor series for $\ln(x)$ or by rewriting it as $\frac{\ln(x)}{\frac{1}{x}}$ and using L'Hopital's.

8. Originally Posted by xalk
Does not the product (ln x)*x ,as x goes to zero go to $-\infty*0$,which is indeterminent,if im not mistaken??
Huh? That is weird, because you already posted the solution yourself

Originally Posted by xalk
$\lim_{x\rightarrow 0^+}({x\ln x}) = \lim_{x\rightarrow 0^+}{\frac{\ln x}{\frac{1}{x}}} = \lim_{x\rightarrow 0^+}{\frac{\frac{1}{x}}{\frac{-1}{x^2}}} = \lim_{x\rightarrow 0^+}({-x}) = 0$

This is exactly what you asked for, and you gave the answer in your post before. Quite funny

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

By considering their graphs it's easy to see that

$

\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:
$
\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

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

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

rearraging we arrive at

$
\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.

10. Another way to view it, is remembering that if $x \rightarrow 0$ $\ln(x) \rightarrow x-1$.

So $\lim_{x \to 0} x^x=\lim_{x \to 0} e^{x \ln (x)}=\lim_{x \to 0} e^{x (x-1)} = 1$

11. Originally Posted by Ruun
Another way to view it, is remembering that if $x \rightarrow 0$ $\ln(x) \rightarrow x-1$.
False.
You would get that $\lim_{x\to 0} \ln(x)=-1$

12. You're right, didn't thought twice. Thanks for the correction!

13. Some years ago I demonstrated in rigorous way that the expression $0^{0}$ is not at all 'undetermined' but it is...

$0^{0}=1$

That meas that ...

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

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

Kind regards

$\chi$ $\sigma$

14. Originally Posted by chisigma
Some years ago I demonstrated in rigorous way that the expression $0^{0}$ is not at all 'undetermined' but it is...

$0^{0}=1$
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 ...

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

... is continous in $x=0$ because the limit and the value of the function are identical...
And the limit as x comes from the left : $x\to 0^-$ ?

15. Originally Posted by Moo
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)
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 $1$, so that...

$0^{0}=1$

Please avoid tedious further discussions ...

Kind regards

$\chi$ $\sigma$

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