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I am interested in the limit

$$ \lim_{n \rightarrow \infty} \sqrt[n]{n^3}$$

Can we simply conclude that: $$ \lim_{n \rightarrow \infty} (\sqrt[n]{n})^3= 1^3=1.$$ I have proven that $\sqrt[n]{n}\rightarrow1$ earlier in this textbook. Also since the limit of a power is the power of the limit.

amWhy
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5 Answers5

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Yes, we can simply do that. Since the exponent $^3$ is a constant neutral number (meaning we may interpret it as a fixed number of multiplications) we can move the limit inside of it. So if you already know $\lim_{n\to\infty}\sqrt[n]n=1$ then that's a full proof.

Arthur
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  • This answer is insufficient. It is not enough to state that the exponent is constant. An essential property is missing. –  Nov 22 '18 at 10:06
  • @YvesDaoust Raising to the power of a constant is continuous around $1$? – Arthur Nov 22 '18 at 10:46
  • This is precisely the missing property. By the way, even a variable exponent would yield a continuous function. –  Nov 22 '18 at 14:37
  • @YvesDaoust Sure you can do it with a variable exponent as well, but then you have to be extra careful. You can't put the limit inside of, say, $\sqrt[n]{n}^n$ and expect to get a sensible answer. You need a constant exponent for that (or at least an exponent converging to a nice value). And I could argue for this without continuity of exponentiation at all. Just using that $\sqrt[n]{n}^3 = \sqrt[n]{n}\cdot\sqrt[n]{n}\cdot\sqrt[n]{n}$, and you can take the limits of those separately (something which is usually proven quite early and taken as implicit). – Arthur Nov 22 '18 at 14:40
  • What I mean is that "since the exponent $^3$ is constant" is not the reason. –  Nov 22 '18 at 15:00
  • @YvesDaoust So you're saying that I specified an unnecessarily strong property of the number $3$, and should've used a weaker one? I disagree. At any rate, I clarified somewhat. Better now? – Arthur Nov 22 '18 at 15:13
  • @YvesDaoust I don't know it. I say $3$ is constant, you say "that's not what you're after, what you're after is that the sequence $(3)n$ _converges". That looks to me like you think I used a too strong property. – Arthur Nov 22 '18 at 16:01
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Yes we are allowed to do that since for continuity

$$\lim_{x\to x_0} f(x)=L\in \mathbb{R} \implies \lim_{x\to x_0} [f(x)]^k=\left[\lim_{x\to x_0} [f(x)\right]^k=L^k$$

Indeed in that particular case since $\sqrt[n]{n}\to 1$

$$\forall \epsilon>0 \quad \exists \bar n \quad \forall n>\bar n \quad |\sqrt[n]{n}-1|<\epsilon$$

we have that, assuming $\sqrt[n]{n}<2$, $\forall \bar \epsilon=7\epsilon>0$

$$|(\sqrt[n]{n})^3-1|=|\sqrt[n]{n}-1|\cdot |\sqrt[n]{n^2}+\sqrt[n]{n}+1|< 7\epsilon=\bar \epsilon \quad \forall n>\bar n$$

and then $(\sqrt[n]{n})^3\to 1$.

user
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  • 1
    You would need this to be an "if and only if" for this to apply here. – Will Sherwood Nov 21 '18 at 21:15
  • @WillSherwood Yes but in that case the asker was interested in that particular implication. Do you think that this is the motivation for the downvoting? Thanks – user Nov 21 '18 at 21:24
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Yes, because the cubic root function is continuous, so that you can swap the limit and the root.

Continuity at $1$ is ensured by the fact that

$$|\sqrt[3]{1+\delta}-1|=\left|\frac{\delta}{(\sqrt[3]{1+\delta})^2+\sqrt[3]{1+\delta}+1}\right|<\frac\delta3<\epsilon$$ holds with $\delta<3\epsilon$.

2

You know that $\sqrt[n]{n} \to 1$.

Also $\sqrt[n]{n^k} =(\sqrt[n]{n})^k $ so

$\begin{array}\\ \sqrt[n]{n^k}-1 &=(\sqrt[n]{n})^k-1\\ &=(\sqrt[n]{n}-1)\sum_{j=0}^{k-1} \sqrt[n]{n}^j\\ \text{so}\\ |\sqrt[n]{n^k}-1| &=|(\sqrt[n]{n}-1)\sum_{j=0}^{k-1} \sqrt[n]{n}^j|\\ &=|(\sqrt[n]{n}-1)||\sum_{j=0}^{k-1} \sqrt[n]{n}^j|\\ &\le|(\sqrt[n]{n}-1)|\sum_{j=0}^{k-1} |\sqrt[n]{n}^j|\\ &\le|(\sqrt[n]{n}-1)|\sum_{j=0}^{k-1} |\sqrt[n]{n}^n|\\ &=|(\sqrt[n]{n}-1)|\sum_{j=0}^{k-1} |n|\\ &=|(\sqrt[n]{n}-1)|kn\\ \end{array} $

Therefore, to make $|\sqrt[n]{n^k}-1| \le \epsilon$, choose $n$ large enough so that $|(\sqrt[n]{n}-1)| \le \frac{\epsilon}{kn} $.

marty cohen
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2

You can if you know four things.

1) If $f$ is continuous and $a_n \to M$ and for all $a_n$ that $f(a_n)$ is defined and $f(M)$ is defined then $a_n\to M$ means $f(a_n) \to f(M)$.

But you DO have to prove this sometime. I'm sure your text has proven that somewhere.

And

2) you need to know that $\sqrt[n]{n} \to 1$.

But you claim you have already shown that.

And 3) that $\sqrt[n]{n^3} = (\sqrt[n]{n})^3$.

Which is basic. It follows that as for all $M > 0$ and $n\in \mathbb N$ there is a unique $k = \sqrt[n]{M}$ so that $k^n = M$. And as $(k^j)^n= (k^n)^j = M^j; for j \in \mathbb N$ it follows that $(\sqrt[n]{M})^j= \sqrt[n]{M^3}$.

And finally you need to know 4) that $()^3: \mathbb R \to \mathbb R$ is continuous.

Which is .... basic. But you should have proven that sometime.

So $\lim \sqrt[n]{n^3} = \lim (\sqrt[n]{n})^3 = (\lim \sqrt[n]{n})^3 = 1^3 =1$.

fleablood
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