Evaluating $\lim\limits_{x \rightarrow0^+} x^{\frac{1}{x}}$

Question

How would I evaluate the following limit?

$$ \lim\limits_{x \rightarrow0^+} x^{\frac{1}{x}} $$

What I've tried

I've tried a few things to evaluate this limit, but I can't seem to fully figure it out. Here's what I tried:

$\text{Let } y = \lim_\limits{x \rightarrow0^+} x^{\frac{1}{x}} \\ \ln y = \ln \lim_\limits{x \rightarrow0^+} x^{\frac{1}{x}} \\ \ln y = \lim_\limits{x \rightarrow0^+} \ln x^{\frac{1}{x}} \\ \ln y = \lim_\limits{x \rightarrow0^+} \frac{1}{x} \ln x \\ \ln y = \lim_\limits{x \rightarrow0^+} \frac{\ln x}{x}$

This gets me to a weird intermediate form of $\frac{-\infty}{0^+}$. This is not a form that I could, for example, use L'Hospital's Rule to evaluate. I've looked some things up, and people say that $\frac{-\infty}{0^+}$ just equals $- \infty$. Why is this true, though? How could this be proven? Regardless, continuing...

$\ln y = - \infty \\ y = e^{-\infty} \\ y = 0\\ \lim\limits_{x \rightarrow0^+} x^{\frac{1}{x}} = 0$

This gives us an answer of $0$, which is the right answer. I just don't understand why $\frac{-\infty}{0^+} = -\infty$

Am I going about it in completely the wrong way? How should I evaluate this limit correctly, maybe by even using L'Hospital's Rule? And why does that previously mentioned intermediate form equal negative infinity?

Answer 1

Addressing why limits of the form $\frac{-\infty}{0^+}$ give $-\infty$: If you take a huge negative number and divide it by a tiny positive number, you get an even-huge-er negative number. Example: -100 / 0.01 = -10000.

Making the numerator more negative just makes the result even more negative. Making the denominator a smaller positive number just makes the result even more negative. Both of these effects are working together, so there's no ambiguity: the result goes to $-\infty$ as the numerator goes to $-\infty$ and the denominator goes to $0$ while staying positive. The fact that the two effects are working together could be restated as saying that $\frac{-\infty}{0^+}$ is a determinate form.

This is as opposed to an indeterminate form, like $\frac{0^+}{0^+}$, where the two effects are working against each other: making the denominator smaller makes the result grow, while making the numerator smaller makes the result shrink. So the answer depends on how quickly the numerator and denominator are shrinking. For example, $\lim_{x\to 0^+} \frac{x^2}{x} = 0$ because the numerator "wins" and is shrinking much faster than the denominator, while $\lim_{x\to0^+} \frac{x}{x^2} = \infty$ because the denominator "wins" and shrinks faster than the numerator.

Answer 2

We can simply observe that eventually $\frac 1 x \ge 1 \iff 0<x\le 1$ then

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

and by squeeze theorem $x^{\frac{1}{x}} \to 0$.

Answer 3

$\lim_ {x \to 0^+} x^{1/x}=?$

$x=1/n$

$\lim_{n \to \infty} (1/n)^n=0$

$x^{1/x}= e^{\frac{\ln x}{x}}$

$\lim_{x \to 0^+} \frac{\ln x}{x}$

$0<x<1 \implies -\infty < \ln x< 0$

$0<x<1 \implies 1<1/x<\infty$

Since the numerator is always negative and the denominator is always positive, the ratio is always negative and its absolute value increases without bound i.e. $ \frac{\ln x}{x} \to -\infty$

This is also the same result you got when you used l'Hopital's Rule.

So the original function goes to $0$.

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Suppose $I:\lim_{x\to 0^+} f(x)\to \infty $ and $II:\lim_{x \to 0^+} g(x) \to -\infty$

Then $\lim_{x \to 0^+} f(x)g(x) \to -\infty$

I: means $\forall M>0, \exists \delta_1>0 $ so that $0<x<\delta_1 \implies f(x)>M$

II: means $\forall N>0, \exists \delta_2>0 $ so that $0<x<\delta_2 \implies g(x)<-N$

Let $\delta = \min(\delta_1, \delta_2)$

Then $0<x<\delta \implies f(x)>M \land g(x)<-N$

$g(x)<-N \implies -g(x)>N$

Multiply the M and N inequalities to get

$-f(x)g(x)>MN\implies f(x)g(x)< -MN$

$M$ and $N$ are both positive so their product is also positive.

This means $f(x)g(x)$ is less than any negative value. If a quantity is less than any negative value, you can say it tends to negative infinity.