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$\displaystyle\lim_{x\to 0}^{}\sin{x}^{x}$. I tried this $ P =\displaystyle \lim_{x\to 0}(\sin{x})^x$ taking the natural log on both side we get $\log_{e}P= \displaystyle \lim_{x\to 0}\log_{e}(\sin{x})^x$

$=\displaystyle \lim_{x\to 0}x\ln(\sin{x})$

$=\displaystyle \lim_{x\to 0}\frac{\ln(\sin{x})}{1\over x}$

this is $\frac{\infty}{\infty}$

I think we should use L'Hospital's rule, but I can't find the answer.

Martin Sleziak
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Prasenjeet Symon
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  • Well what is $\lim_{x \to 0} x^x$? – mopy May 05 '15 at 16:47
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    $;x^x;$ doesn't exist for $;x<0;$ and thus the wanted limit cannot exist. If you want to talk of the right sided then something can be done...perhaps. – Timbuc May 05 '15 at 16:48
  • If you mean $\lim_{x\to 0}\sin (x^x)$, then $\sin $ being continuous implies $\lim_{x\to 0^+}\sin x^x=\sin\lim_{x\to 0^+} x^x=\sin 1$. But I see that's not what you meant. $\lim_{x\to 0}\sin x^x$ is ambiguous. – user26486 May 05 '15 at 16:59
  • @user31415 that is not the question – RE60K May 05 '15 at 17:00
  • @ADG I know. I just pointed out that $\displaystyle\lim_{x\to 0}^{}\sin{x}^{x}$ is ambiguous, though later statements remove ambiguity. – user26486 May 05 '15 at 17:01
  • @Timbuc your this example is wrong too. – RE60K May 05 '15 at 17:18
  • @ADG I'm not sure what you're referring to: are you saying $;x^x;$ can be well defined as a real function in some left (open) neighbourhood of zero? And what example of mine do you mean? Thank you. – Timbuc May 05 '15 at 17:21
  • @Timbuc yes it doesn't exist for <0, but the limit does. – RE60K May 05 '15 at 17:22
  • @ADG That's impossible,, as per definition, for $;\lim_{x\to x_0} f(x);$ to have the slightest chance to exists it must be true that $;f;$ is defined in some (punctured) neighbourhood of $;x_0;$ , and that means both-sided neighbourhood. Also, in your answer, could you please explain what does $;\log\sin x;$ mean for $;\sin x<0;$ ?! – Timbuc May 05 '15 at 17:24
  • @Timbuc if you're not getting it see this – RE60K May 05 '15 at 17:27
  • @ADG I can't agree with that, and as far as I can see neither do several members here and, possibly, some authors here and there. Anyway, even if that convention is adopted, I think a little note explaining that would be in place. – Timbuc May 05 '15 at 17:28
  • @Timbuc well I can't say beyond that since I can't make you agree. I don't wish to argue. – RE60K May 05 '15 at 17:36
  • @ADG In Hijab's "Int. to Calculus and Real Analysis", in Thomas-Finney's " Calculus and Anal. Geom.", in Stewart's "Calculus, Early Trascendentals", etc. it is explicitly required the function to be defined in an open neighborhood of $;x_0;$ , except perhaps at that point itself, for the limit (without sides) to be existent. But I guess it all is a matter of definition... – Timbuc May 05 '15 at 17:38
  • re: I don't wish to argue. Opinions. – RE60K May 05 '15 at 17:42

2 Answers2

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For the right-hand side limit:

Suppose $0<x<\pi$. Then: $$x\ln(\sin x)=\frac x{\sin x}\cdot \sin x \ln(\sin x)$$ Now $\dfrac x{\sin x}\to 1$ as $x\to 0$, and $\,u\ln u\to 0$ as $u\to 0_+$, so
$$x\ln(\sin x)\to 0\enspace\text{and}\quad\lim_{x\to 0_+} (\sin x)^x = 1$$

Bernard
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$$\lim_{x\to0}(\sin x)^x=\lim_{x\to0}e^{x\ln(\sin x)}=\lim_{x\to0}e^{\displaystyle \frac{\ln(\sin x)}{1/x}}\stackrel{L'H}=\lim_{x\to0}e^{\displaystyle \frac{\cot x}{-1/x^2}}=\lim_{x\to0}e^{\displaystyle -x\times\frac{x}{\tan x}}=e^{0\times1}=1$$

RE60K
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  • This can't be right as $;x<0\implies (\sin x)^x;$ isn't defined (for example, try with $;x=-\frac1{20};$ ...) – Timbuc May 05 '15 at 17:00
  • @Timbuc well you're having it wrong, sorry but please read about limits again. – RE60K May 05 '15 at 17:17
  • First, I think the function in the limit is actually $;\sin\left(x^x\right);$ and not what you wrote, but whatever: How can a negative number be raised to a power $;n/m;$ , with $;n;$ an odd natural and $;m;$ and even one? Maybe I'm missing something here... – Timbuc May 05 '15 at 17:19
  • First, see this and yes you are missing something [I'm just avoiding it ATM so that you find it yourself or else, if you need, I may tell you] – RE60K May 05 '15 at 17:22
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    Sorry, I can't see your point, and I don't know what ATM stands for. And please be kind to me and tell me exactly what it is I'm missing instead of repeating over and over that I'm missing something. Thanks. – Timbuc May 05 '15 at 17:27
  • @Timbuc nevermind, I never intended to do that. I just told you to see original post by the OP and I've linked you to this where it is explained that lim can exist for undefined functions on one side. – RE60K May 05 '15 at 17:28