My limit challenge

[lookagain]

Given that \(\displaystyle 0^0 \ \ is \ \ indeterminate, \ \ and \ \ that \ \ \displaystyle\lim_{x\to 0^+} x^x \ = \ 1, \ \ and\)

using your algebra and/or calculus skills, determine the value of:


\(\displaystyle \displaystyle\lim_{x\to0^+}\bigg(1 - x^x\bigg)^{(1 - x^x)} \)

 

[MarkFL]

Assuming the limit exists, I would state:

\(\displaystyle \displaystyle \lim_{x\to0^{+}}\left(1-x^x \right)^{1-x^x}=L\)

Take the natural log of both sides:

\(\displaystyle \displaystyle \ln\left(\lim_{x\to0^{+}}\left(1-x^x \right)^{1-x^x} \right)=\ln(L)\)

\(\displaystyle \displaystyle \lim_{x\to0^{+}}\ln\left(\left(1-x^x \right)^{1-x^x} \right)=\ln(L)\)

\(\displaystyle \displaystyle \lim_{x\to0^{+}}(1-x^x)\ln\left(1-x^x \right)=\ln(L)\)

\(\displaystyle \displaystyle \lim_{x\to0^{+}}\dfrac{\ln\left(1-x^x \right)}{\dfrac{1}{1-x^x}}=\ln(L)\)

We now have the indeterminate form \(\displaystyle \dfrac{\infty}{\infty}\), so application of L'Hôpital's rule gives:

\(\displaystyle \displaystyle \lim_{x\to0^{+}}\dfrac{\dfrac{\dfrac{d}{dx}(1-x^x)}{1-x^x}}{-\dfrac{\dfrac{d}{dx}(1-x^x)}{(1-x^x)^2}}=\ln(L)\)

\(\displaystyle \displaystyle \lim_{x\to0^{+}}(x^x-1)=\ln(L)\)

\(\displaystyle 0=\ln(L)\)

\(\displaystyle L=e^0=1\)

 

[Deleted member 4993]

And it can be confirmed by plotting (@wolframalpha.com)

 

[lookagain]

\(\displaystyle Let \ \ 0 < x < 1.\)

Can I accurately state \(\displaystyle 0^x < x^x < 1^x \ ?\)

If so, then \(\displaystyle 0 < x^x < 1 \ \ for \ \ 0 < x < 1.\)

Then \(\displaystyle (-1)(0) > -1(x^x) > -1(1).\)

\(\displaystyle 0 > -x^x > -1\)

\(\displaystyle 0 + 1 > -x^x + 1 > -1 + 1\)

\(\displaystyle 1 > 1 - x^x > 0\)

\(\displaystyle Or, \ \ 0 < 1 - x^x < 1\)

Then \(\displaystyle 1 - x^x \ = \ \epsilon, \ \ where \ \ \ 0 < \epsilon < 1.\)

Then \(\displaystyle \epsilon \ \ approaches \ \ 0^+ \ \ as \ \ x \ \ approaches \ \ 0^+.\)

The original expression becomes:

\(\displaystyle \lim_{\epsilon\to 0^+}{\epsilon^{\epsilon}} \ = \ 1\)


(This is a consequence of the second given in the first line of my original post.)

 

[HallsofIvy]

\(\displaystyle Let \ \ 0 < x < 1.\)

Can I accurately state \(\displaystyle 0^x < x^x < 1^x \ ?\)

Yes, as long as 0< x< 1, x to any power is positive and less than 1.

If so, then \(\displaystyle 0 < x^x < 1 \ \ for \ \ 0 < x < 1.\)

Then \(\displaystyle (-1)(0) > -1(x^x) > -1(1).\)

\(\displaystyle 0 > -x^x > -1\)

\(\displaystyle 0 + 1 > -x^x + 1 > -1 + 1\)

\(\displaystyle 1 > 1 - x^x > 0\)

\(\displaystyle Or, \ \ 0 < 1 - x^x < 1\)

Then \(\displaystyle 1 - x^x \ = \ \epsilon, \ \ where \ \ \ 0 < \epsilon < 1.\)

Then \(\displaystyle \epsilon \ \ approaches \ \ 0^+ \ \ as \ \ x \ \ approaches \ \ 0^+.\)

The original expression becomes:

\(\displaystyle \lim_{\epsilon\to 0^+}{\epsilon^{\epsilon}} \ = \ 1\)


(This is a consequence of the second given in the first line of my original post.)