1
$\begingroup$

I was wondering if i could get any help with the following:

$$ \lim_{x \rightarrow 0^{+}} \frac{x^x -1}{x} $$.

thank you.

My attempt:

$$ \lim = \frac{e^{x \ln(x)} - 1}{x} = \lim ( x \ln (x)( x \ln(x) + 1)) $$

$\endgroup$
  • 1
    $\begingroup$ Your $\frac{x\ln x-1}{x}$ should have been $\frac{e^{x\ln x}-1}{x}$. And your differentiation of $x\ln x$ was not correct. $\endgroup$ – André Nicolas Sep 26 '15 at 18:18
  • $\begingroup$ @AndréNicolas thank you $\endgroup$ – user2804865 Sep 26 '15 at 19:04
  • $\begingroup$ You are welcome. The basic L'Hospital's Rule strategy was fine. $\endgroup$ – André Nicolas Sep 26 '15 at 19:06
5
$\begingroup$

Use L'Hopital's rule to get:

$$\lim_{x \to 0} x^x(\ln x+1)=\lim_{x \to 0}x^x \ln x-\lim_{x \to 0} x^x=(\lim_{x \to 0} x^x \ln x) -1=((\lim_{x \to 0} x^x)(\lim_{x \to 0} \ln x))-1=(1)(-\infty)-1=-\infty$$

The limit does not exist.

$\endgroup$
4
$\begingroup$

We may also avoid De l'Hopital theorem. Since $$ \lim_{t\to 0}\frac{e^t-1}{t}=1\quad\text{and}\quad \lim_{x\to 0^+} x\log(x)=0,$$ we have: $$ \lim_{x\to 0^+}\frac{x^x-1}{x}=\lim_{x\to 0^+}\frac{e^{x\log x}-1}{x\log x}\cdot\log(x) = \lim_{x\to 0^+}\log(x) = -\infty.$$

$\endgroup$
  • $\begingroup$ Nice. Does it make a difference to use ln or log in this case? $\endgroup$ – GnP Sep 26 '15 at 18:32
  • $\begingroup$ @gnp: no, it doesn't. For my standards, $\log$ means the natural logarithm as well as $\ln$. $\endgroup$ – Jack D'Aurizio Sep 26 '15 at 18:35
-1
$\begingroup$

Use an asymptotic development:

$x^x=\mathrm e^{x\ln x}=1+x\ln x+o(x\ln x)$, hence $$\frac{x^x-1}x=\frac{x\ln x+o(x\ln x)}x=\ln x+o(\ln x)\sim\ln x\xrightarrow[x\to 0^+]{}-\infty.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.