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

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

Thanks

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    $\begingroup$ There were notions doing still more damage than "indeterminate form". But not very many. $\endgroup$
    – user436658
    Dec 28, 2020 at 14:08
  • $\begingroup$ No, google cauchy sequences, in R with the standard metric, converges. $\endgroup$
    – yugikaiba
    Dec 28, 2020 at 14:18
  • $\begingroup$ The base is fixed , so this is not an indeterminate expression. $\endgroup$
    – Peter
    Dec 28, 2020 at 14:20

4 Answers 4

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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

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

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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.


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.
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    $\begingroup$ This is a very formal approach, and definitely correct, but I fear with the way the question was phrased, will be relatively useless for the OP who seems to have no background in formal thinking. Still +1 for others who may come here to find a more formal approach $\endgroup$
    – gt6989b
    Dec 28, 2020 at 14:14
  • $\begingroup$ @gt6989b My intent is to show how the formal approach leaves no room for interpretation, and thus motivate the OP to start thinking more formally. In mathematics, in my view, the formal approach is the only way of thoroughly understanding what is happening when we say anything. There's a reason every math student gets the formal approach beaten into them in their first years of study. $\endgroup$
    – 5xum
    Dec 28, 2020 at 14:20
  • $\begingroup$ I agree with you in principle, but after doing this for a while, I don't know anyone who goes through the formal argument when they need to take a limit. Formalization is needed to convince yourself that your intuition is true, and without it, the end result is hanging in the air unsupported by anything... but in my experience, building intuition on formalisms does not work very well. You have to start from the definition and explain what each piece of this means "in human language" - then do a lot of problems and then hope that builds enough intuition for them to go back and forth easily... $\endgroup$
    – gt6989b
    Dec 28, 2020 at 14:36
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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).

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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.


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 $1$.

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