# Finding $\lim_{x \to 0^{+}} x^{\frac{1}{x}}$ by L'Hôpital's Rule.

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

• Note $\infty \cdot (-\infty)$ is not an indeterminate form. Prove this: if $f(x) \to \infty$ and $g(x) \to -\infty$, then $f(x)g(x) \to -\infty$. Commented Jul 7, 2023 at 20:58
• How your graphing calculator says that in the original form of the exercise, the limit $= 1$? It should be $0$.
– user
Commented Jul 7, 2023 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. Commented Jul 8, 2023 at 1:08
• George Marsaglia published on usenet aged 86. And then he suddenly stopped. Commented Jul 8, 2023 at 5:51

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

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

• Thanks for the answer. Much appreciated. Commented Jul 8, 2023 at 1:09
• 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
Commented Jul 8, 2023 at 4:48
• 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. Commented Jul 8, 2023 at 5:55

Let $$M=\frac{1}{x}$$. Then $$\lim_{x\rightarrow 0^{+}}\large{x^{\frac1x}}=\lim_{M\rightarrow\infty}\frac{1}{M^M}=0 .$$

• Very nice way too!
– user
Commented Jul 8, 2023 at 6:00