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:
$$ \lim = \frac{e^{x \ln(x)} - 1}{x} = \lim ( x \ln (x)( x \ln(x) + 1)) $$
Asked
Active
Viewed 177 times
1
-
1Your $\frac{x\ln x-1}{x}$ should have been $\frac{e^{x\ln x}-1}{x}$. And your differentiation of $x\ln x$ was not correct. – André Nicolas Sep 26 '15 at 18:18
-
@AndréNicolas thank you – user2804865 Sep 26 '15 at 19:04
-
You are welcome. The basic L'Hospital's Rule strategy was fine. – André Nicolas Sep 26 '15 at 19:06
3 Answers
5
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.
Aleksandar
- 1,771
-
Why does it not exist? How would you show one side goes to something else. – Ahmed S. Attaalla Sep 26 '15 at 18:22
-
-
I thought DNE was only used to show that the RH limit does not agree with the LH limit. – Ahmed S. Attaalla Sep 26 '15 at 18:26
-
1
-
1@AhmedS.Attaalla http://math.stackexchange.com/questions/127689/why-does-a-limit-at-infinity-not-exist and http://math.stackexchange.com/questions/3203/infinite-limits – Aleksandar Sep 26 '15 at 18:29
4
We may also avoid De L'Hospital 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: $$ \lim_{x\to 0^+}\frac{x^x-1}{x}=\lim_{x\to 0^+}\frac{e^{x\log x}-1}{x\log x}\cdot\log(x) = \lim_{x\to 0^+}\log(x) = -\infty.$$
Bernard
- 175,478
Jack D'Aurizio
- 353,855
-
-
@gnp: no, it doesn't. For my standards, $\log$ means the natural logarithm as well as $\ln$. – Jack D'Aurizio Sep 26 '15 at 18:35
0
Use an asymptotic expansion:
$x^x=\mathrm e^{x\ln x}=1+x\ln x+o(x\ln x)$, hence $$\frac{x^x-1}x=\frac{x\ln x+o(x\ln x)}x=\ln x+o(\ln x)\sim\ln x\xrightarrow[x\to 0^+]{}-\infty.$$
Bernard
- 175,478