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I need someone to tell me the step I'm missing or doing incorrectly. The problem is:

$$\lim _{ x\rightarrow 0 }{ { x }^{ x } } $$

$1$. $\ln x^x$

$2$. $x\ln { x } $

$3$. This is the step I don't follow: $\frac { \ln { x } }{ \frac { 1 }{ x } } $

I come up with: $\frac { \ln { x } }{ x } $ at this point.

I'm missing this step in all the natural $\log$/l'hopitals problems I work. I picked the most simple for ease of explanation, if someone could be so kind?

haqnatural
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Tonantzin
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3 Answers3

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$$\lim _{ x\rightarrow 0 }{ { x }^{ x }= } \lim _{ x\rightarrow 0 }{ { e }^{ x\ln { x } }= } \lim _{ x\rightarrow 0 }{ { e }^{ \frac { \ln { x } }{ \frac { 1 }{ x } } }\overset { L'h\quad rule }{ = } } \lim _{ x\rightarrow 0 }{ { e }^{ \frac { \frac { 1 }{ x } }{ -\frac { 1 }{ { x }^{ 2 } } } }= } \lim _{ x\rightarrow 0 }{ { e }^{ -x }=1 } $$

haqnatural
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  • That is cool. I did not realize where 'e' fit into this. When x becomes 1/x, is that just a solid rule for x or whatever number comes before ln in this situation with 'e', that they are moved to the denominator in the form of 1/n, or is there some other methodology that leads to that step? I've not got a great background with logarithms or 'e', so this seems out of the blue to me. Thank you! – Tonantzin Jul 31 '16 at 00:26
  • @Tonantzin,note that we know of from the property of log. that ${ a }^{ \ln { b } }={ b }^{ \ln { a } }$ in this case we have $a=e,b={ x }^{ x }$,hope it will help.you're welcome – haqnatural Jul 31 '16 at 00:30
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Even without L'Hospital, considering $$A=x^x \implies \log(A)=x \log(x)\implies \log(A)\to 0\implies A \to 1$$

Claude Leibovici
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  • Except that it is not entirely obvious that $\lim_{x\to0+}x\log x=0$ when $\log x\to-\infty$. I'm sure you know the argument, but given that this more or less equivalent to the end claim... – Jyrki Lahtonen Jul 31 '16 at 05:37
  • I cannot do anything else than totally agreeing with you ! Cheers. – Claude Leibovici Jul 31 '16 at 05:41
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I interpret your question to be about how $$x\ln x =\frac{\ln x}{\frac1x}$$

Here, they're using the rule that $\displaystyle x = \frac1{\frac1x}$ and that $\frac1b \cdot a = \frac ab$

That means that, with a bit more steps, we have $$x\cdot \ln x = \frac1{\frac1x}\cdot \ln x = \frac{\ln x}{\frac1x}$$

The reason this is useful, is that the limit now gives an indeterminate form that qualifies for L'Hopital.

Alec
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