# Why $0^{\infty}$ isn't indeterminate form?

I was trying to calculate $$\lim _{x \rightarrow 0} x^{\frac{1}{x}}$$ I know left hand limit is not equal to right hand limit, hence limit doesn't exist. But I was trying to get their values as well. Then I came to the question why $$0^{\infty}$$ isn't an indeterminate form ?

• An expression being an indeterminate form does not mean the limit does not exist. In fact, the right limit of $x^{1/x}$ is $0$ and the left limit is undefined in $\mathbb{R}$ (what is $(-1/\pi)^{-\pi}$?) Commented Apr 30, 2023 at 5:19
• The function isn't even defined for $x<0.$ Wat is $(-2/3)^{-3/2}?$ Commented Apr 30, 2023 at 5:21
• Limits of kind $(+0)^{+\infty}=0$, $(+0)^{-\infty}=+\infty$. Both forms are not indeterminate. Commented Apr 30, 2023 at 7:34
• The left side limit is not just undefined, the function is undefined on the left side. There is nothing to take the limit of. The left side is not a place where the function is defined,.. Usually, when we say the "limit does not exist" it is because there is something to take the limit of. There isn't here. Commented Apr 30, 2023 at 7:34
• In particular, since the function to the left is not defined, it is not of the "form" $0^\infty,$ and thus this doesn't contradict the statement that $0^\infty$ is determinate. Commented Apr 30, 2023 at 7:41

The left hand limit does not exist because the function is not defined on the left side. So the right side limit is of the "form" $$0^\infty,$$ but the left side limit is not.

So this does not prove that $$0^\infty$$ is indeterminate.

As a general rule, we might say $$\lim_{x\to0}\sqrt x=0,$$ but the danger is a case like $$\lim_{x\to a}\sqrt{f(x)},$$ where we don't know where $$f(x)$$ is positive or negative. It gets very complicated to define something like: $$\lim_{x\to0}x\sqrt{\sin\frac1x}$$ where there are undefined values all around $$0.$$

In general, if $$X$$ is the domain of a function - the set of values $$x$$ where $$f(x)$$ is defined - we restrict the definition of limit to be only for $$x$$ in the domain. But for $$\lim_{x\to a} f(x)$$ to be defined, we need to require that $$a$$ is a "limit point" of the set $$X.$$

For example, talking about $$\lim_{x\to0}\sqrt{x-1}$$ is meaningless, because the function isn't defined when $$|x|<1.$$

Indeed, the "right side" limit of a function, often written $$x\to a^+,$$ can be thought of as asking what the limit is if we restrict the function to the domain $$X=\{x\mid x>a\},$$ and similar for left side limits and $$X=\{x\mid x

The base eventually becomes less than 1 so when you raise it to a high power it becomes even closer to 0, so the limit is 0. So there is no ambiguity with the limits that calls for indeterminate forms.

More precisely, let $$f(x)$$ and $$g(x)$$ be continuous functions, and suppose that $$\lim_{x \to c}f(x)=0$$ and $$\lim_{x \to c}g(x)=\infty$$. Also assume that $$f(x)>0$$ for all $$x$$. I claim that $$\lim_{x \to c}f(x)^{g(x)}=0$$. Let $$\epsilon>0$$ and assume that $$\epsilon <1$$. We need to show that there is some $$\delta>0$$ such that $$|f(x)^{g(x)}|< \epsilon$$. There is some $$\delta_1>0$$ such that for all $$x \in (c-\delta_1,c+\delta_1)$$, $$|f(x)|<\epsilon$$ and there is some $$\delta_2>0$$ such that for all $$x \in (c-\delta_2,c+\delta_2)$$, $$g(x)>2$$. Now let $$\delta<\min\{\delta_1,\delta_2\}$$. When $$x \in (c-\delta,c+\delta)$$, $$|f(x)^{g(x)}|=|f(x)|^{g(x)}<\epsilon^2 < \epsilon$$.This shows that $$\lim_{x \to c}f(x)^{g(x)}=0$$.

• You need $f$ to be a positive function for $f(x)^{g(x)}$ to be defined, which is the source of the OP is having, OP wonders why the limits from both sides don't agree when $0^\infty$ is, they are told, not an indeterminate form. The problem is that there is no function on left. Commented Apr 30, 2023 at 7:49
• I added the condition. Anyway, my post answers the question of why $0^\infty$ isn't an indeterminate form. OP does seem to believe that if the limit doesn't exist then somehow that has something to do with indeterminate forms. Commented Apr 30, 2023 at 8:51

In general for $$f(x)>0$$

$$\lim _{x \rightarrow x_0} \left(f(x)\right)^{g(x)}$$

with $$\lim _{x \rightarrow x_0} f(x)=0$$ and $$\lim _{x \rightarrow x_0} g(x)=\infty$$ is not an indeterminate form because

$$\left(f(x)\right)^{g(x)} =e^{g(x)\log (f(x))} \to e^{-\infty}=0$$

As noticed, in this particular case, the limit is well defined only for $$x>0$$ that is

$$\lim _{x \rightarrow 0^+} x^{\frac{1}{x}}=\lim _{x \rightarrow 0^+} e^{\frac{\log x}x}=0$$ since $$\frac{\log x}x =\frac1x \cdot \log x \to -\infty$$.

Refer also to

if $$y = x ^{1/x}$$ then $$\log y = \log x /x$$. Meaning that your limit is $$\exp( \lim _{x\to 0} \log x / x)$$ which indeed is $$\pm \infty/0$$ indeterminate.

• by the way limit DNE as commentors above pointed out the function is not defined for $x < 0$. Commented Apr 30, 2023 at 5:24
• No, it is of the form $-\infty/0,$ or $-\infty\cdot\infty.$ Commented Apr 30, 2023 at 7:45
• There is definition of limit which says limit exists in this case ($S=(0;+\infty)$): en.wikipedia.org/wiki/Limit_of_a_function#More_general_subsets Commented Apr 30, 2023 at 7:49