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I want to compute the limit of $|x|^{\frac{1}{x}}$ when $x$ tends to $0$.

When $x$ tends to $0$ from above, I get ${(0^+)}^{\frac{1}{0^+}}= {(0^+)}^{+\infty}=0$. However when, I approach $0$ from below I get $ {(0^+)}^{\frac{1}{0^-}} = {(0^+)}^{-\infty}$ and I can't give a value for this. ( why ?)

However, I was able to find a value for this limit by going through the exponential definition of $x$ as follow : $$\lim_{x \to 0^-} |x|^{\frac{1}{x}} =\lim_{x \to 0^-} e^{\frac{1}{x} \ln(|x|)}=\lim_{x \to 0^-} e^{\frac{\ln(|x|)}{x} } = e^{\lim_{x \to 0^-}\frac{\ln(|x|)}{x} }= e^{\frac{-\infty}{0^-}}=e^{+\infty}=+\infty$$ So, my question is, why am I forced to go through this definition? I might have been able to give a value to ${(0^+)}^{- \infty}$

ZchGarinch
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    Informally, $(0^+)^{-\infty} = \frac{1}{(0^+)^{\infty} } = \frac{1}{0^+} = \infty$. – Milten Oct 27 '21 at 18:47
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    "and I can't give a value for this" To be clear, in standard real analysis and the usual real numbers... formally we never write $(0^+)^{+\infty}$ nor do we write $(0^+)^{-\infty}$. If we wanted to write related expressions we would have formalized them with limits. The expressions you write are informal, involving the usage of $\infty$ as though it were a number when it is not. Similarly, your expression involves division by zero which should be avoided. – JMoravitz Oct 27 '21 at 18:47
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    That all being said, however you try to formalize it, the loose idea of "a small positive number" raised to "a large negative power" will result in a large positive number... so if you wanted to "give a value" to $(0^+)^{-\infty}$ it would of course be $+\infty$ so your saying "I can't give a value for this" is short-sighted or misguided. – JMoravitz Oct 27 '21 at 18:48
  • Thank you for your answer which is understandable at my level. I would however like to know if could you show me how to formalize my expressions please? – ZchGarinch Oct 27 '21 at 18:56
  • See https://math.stackexchange.com/questions/2671819/what-do-indeterminate-forms-mean/2671895#2671895 – Michael Hoppe Oct 27 '21 at 19:09
  • In this case what about ${(0^-)}^{+\infty}$ ? I dont see why it's also equal to $0$ ! – ZchGarinch Oct 27 '21 at 21:25
  • @ZchGarinch $(0^-)^\infty$ is still "something small to a large power". Large powers make small numbers smaller, i.e. $(-0.1)^6=0.000001$. So it goes to zero. But! In the case of $(0^-)^\infty$, things might not be well-defined, for example $(-1)^\frac12$ would be a problem (unless we go into a whole nother discussion). But as long as the exponent is always an integer, there is no problem. – Milten Oct 27 '21 at 22:07
  • @MichaelHoppe These cases aren't indeterminate though. – Milten Oct 27 '21 at 22:12

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"show me how to formalize my expressions please". Maybe you mean like this?

Assume $a_n\to 0^+$ and $b_n\to -\infty$ for $n\to \infty$. Note that $a_n>0$ for large enough $n$, so $a_n^{b_n}$ is well-defined large enough $n$. Then $a_n^{b_n}\to \infty$. And you basically did a proof yourself, starting with $$ a_n^{b_n} = e^{b_n\log a_n}. $$ Now, $\log a_n \to -\infty$ by continuity, so $b_n\log a_n\to+\infty$, which gives $$ a_n^{b_n} = e^{b_n\log a_n} \to +\infty. $$ In short, $(0^+)^{-\infty} = +\infty$.

Prerequisites for this proof are $ (-\infty)\cdot(-\infty) = +\infty$ and $c^{+\infty} = +\infty, $ where $c>1$ is a constant greater than $1$.

Milten
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