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

Question

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 ?

Answer 1

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 $X=\{x\mid x<a\}.$

Answer 2

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 $\lim_{x \to c}f(x)^{g(x)}=0$.

Answer 3

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.

Answer 4

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

• Is $0^\infty$ indeterminate?
• How to prove that a form like $0^\infty$ is not an indeterminate form? What makes a form indeterminate?