Isn't $\lim\limits_{x \to 0}1^{1/x}$ be indeterminate

Question

Isn't $\lim\limits_{x \to 0}1^{1/x}$ be indeterminate

Since $1^{\infty}$ is indeterminate.

Thanks

Answer 1

No, here we have

$$(\forall x\ne 0)\;\; f(x)=1^{1/x}=1$$

thus $ f $ is constant at $ (-\infty,0)\cup(0,+\infty)$ and

$$\lim_{x\to 0}f(x)=1$$

Answer 2

No. The limit is equal to $1$. This is because for every $x\in\mathbb R\setminus \{0\}$, you have the equality $1^{\frac1x} = 1$.

There is no such thing as "$1^\infty$", and you will do yourself a massive favor if you stop thinking in sloppy terms and start thinking in definitions.

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The definition you need here is the following:

If $f$ is a function defined on $D_f$, then $$L = \lim_{x\to a} f(x)$$ if and only if for every $\epsilon > 0$, there exists some $\delta > 0$ such that for all $x\in D_f\cap((a-\delta, a)\cup(a, a+\delta))$, the equality $|f(x)-L|<\epsilon$ is true.

This definition is all you need. You don't need vague concepts like "$1^\infty$ or anything like that. In fact, you don't need to think about the concept of infinity at all to prove that the limit of your function is $1$.

You can prove that the limit is $1$ in simple steps:

1. Let $\epsilon > 0$ be arbitrary.
2. Define $\delta = 1$.
3. Take any $x\in (-1, 0)\cup (0, 1)$.
4. Evaluate $|f(x)-L|$, which is straightforward.
5. Conclude that $|f(x)-L|<\epsilon$ is true.
6. Because $\epsilon$ was arbitrary, the claim is true for all epsilon values, so the proof is complete.

Answer 3

You're free to consider it an indeterminate form, if you so wish.

There is no mathematical content in the phrase “indeterminate form”: it's just a way to say that the particular limit may need additional work to be computed and no “direct substitution” is possible. Which is mainly done for the benefit of students, because sometimes they're too fast in jumping to conclusions.

^{Whether it actually benefits students or not is debatable. It's probably a question about how the notion of “indeterminate form” is presented.}

Bear in mind that “indeterminate form” doesn't mean “not computable”. So whether you deem $\lim_{x\to0}1^{1/x}$ to be in indeterminate form has no real consequence. The limit is $1$ nonetheless (because it's the limit of a constant function, by the way).

Answer 4

A limit is never indeterminiate. In can be undefined or it may perhaps not exist. The form $1^\infty$ is said to be indeterminate because from $a_n\to 1$ and $b_n\to+\infty$ alone, we cannot determine tha limit (if it exists at all) of $a_n^{b_n}$ as $n\to \infty$. This doesn't mean that the limit never exists.

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Similarly, the form $0^0$ is said to be indeterminate because $a_n\to 0$ and $b_n\to 0$ allow various behaviours for $a_n^{b_n}$ -- even though the expression $0^0$ is evaluates to $1$.