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I am a 76 years old retiree who loves to read math textbooks. Yeah, crazy, right? The current textbook I am using is Larson's Calculus, 12e. Section 5.6, exercise 50 is stumping me: $$\lim_{x \to 0^{+}} x^{\frac{1}{x}}$$ The thrust of this section is to learn how to use L'Hôpital's Rule, particularly to manipulate a limit in indeterminate form, like this one, ($0^{\infty}$), into a suitable form. Since this exercise has a moving exponent, I've used Logarithmic Differentiation: $$\ln y = \lim_{x \to 0^{+}} \left(\frac{1}{x} \cdot \ln x\right) = \infty \cdot \left(-\infty\right)$$ and again: $$= \lim_{x \to 0^{+}} \frac{\ln x}{x} = \frac{-\infty}{0}$$ But here, I don't know what to do. My graphing calculator says that in the original form of the exercise, the limit = 1. So, I figure I have to come up with a way to get to: $$\ln y = 0$$ $$y = e^{0} = 1$$ Any help is appreciated. Thanks,

Jose

user
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JGonzalezGUS
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    Note $\infty \cdot (-\infty)$ is not an indeterminate form. Prove this: if $f(x) \to \infty$ and $g(x) \to -\infty$, then $f(x)g(x) \to -\infty$. – GEdgar Jul 07 '23 at 20:58
  • How your graphing calculator says that in the original form of the exercise, the limit $= 1$? It should be $0$. – user Jul 07 '23 at 21:12
  • Thanks Edgar and user. The answer from my graphing calculator was for another exercise and had tried just before and I forgot to clear it. My apologies. With the correct input by calculator gives 0, as expected. Thanks again. – JGonzalezGUS Jul 08 '23 at 01:08
  • George Marsaglia published on usenet aged 86. And then he suddenly stopped. – gnasher729 Jul 08 '23 at 05:51

2 Answers2

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We don't need l'Hospital, indeed the original limit leads to (simbolically)

$$ x^{\frac{1}{x}} = (0^+)^\infty$$

which is not an indeterminate form since when the base is small a larger exponent leads to a smaller quantity, for this reason

$$ x^{\frac{1}{x}} \to 0$$

or, as you noticed using logarithm

$$\ln y = \lim_{x \to 0^{+}} \left(\frac{1}{x} \cdot \ln x\right) = \infty \cdot \left(-\infty\right) =-\infty \implies y=e^{\ln y}\to 0 $$

user
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  • Thanks for the answer. Much appreciated. – JGonzalezGUS Jul 08 '23 at 01:09
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    You are welcome! To learn limit properly I suggest, in general, to avoid l’Hospital as a first tool. For indeterminate form we should use standard limit at first, then when necessary Taylor expansion is the most powerful method. In some cases also l’Hospital is fine, but not in general. – user Jul 08 '23 at 04:48
  • You need to create a limit of the form 0/0 or inf/inf before you can use l’Hospital. Here you get exp (-inf * +inf) = exp (-inf) = 0. – gnasher729 Jul 08 '23 at 05:55
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Let $M=\frac{1}{x}$. Then $$\lim_{x\rightarrow 0^{+}}\large{x^{\frac1x}}=\lim_{M\rightarrow\infty}\frac{1}{M^M}=0 .$$

Bob Dobbs
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