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The following is a problem from chapter 5, "Limits", of Spivak's Calculus

  1. (a) For $c>1$, show that $c^{1/n}=\sqrt[n]{c}$ approaches $1$ as $n$ becomes very large. Hint: Show that for any $\epsilon>0$ we cannot have $c^{1/n}>1+\epsilon$ for large $n$.

(b) More generally, if $c>0$, then $c^{1/n}$ approaches $1$ as $n$ becomes very large.

Note that the concept of a sequence hasn't been introduced at this point in the book. Only the basic definition of a limit of a function. I am looking for solutions that only use the latter concept.

My question regards the part $(b)$.

Here is the solution to part $(a)$.

Assume $c>1$ and $f(n)=c^{1/n}$

Bernoulli's Inequality says that

$$\forall \epsilon>0\ \forall n>0,\ 1<1+n\epsilon\leq (1+\epsilon)^n$$

Therefore

$$\lim\limits_{n\to \infty}(1+n\epsilon)=\infty\implies \lim\limits_{n\to\infty}(1+\epsilon)^n=\infty$$

This means that

$$\forall \epsilon>0\ \forall M>0\ \exists N>0\ \forall n, n>N \implies (1+\epsilon)^n>M$$

If we specify $M=c$ then

$$\forall \epsilon>0\ \exists N>0\ \forall n, n>N \implies (1+\epsilon)^n>c \implies 0<c^{1/n}-1<\epsilon$$

Now part $(b)$

Let $c>0$.

Assume $c^{1/n}>1+\epsilon$, for all $n\in\mathbb{N}$.

Then $c>(1+\epsilon)^n>1+n\epsilon$ by Bernoulli's Inequality.

But then $n<\frac{c-1}{\epsilon}$ for all $n \in \mathbb{N}$. But we can choose any natural number $n_1>\left \lfloor \frac{c-1}{\epsilon} \right \rfloor +1$ and have $n_1>\frac{c-1}{\epsilon}$.

Therefore, we have a contradiction, and can infer that

$$\forall \epsilon>0\ \exists M\ \forall n, n>M \implies 0<c^{1/n}<1+\epsilon$$

$$\implies -1<c^{1/n}-1<\epsilon\tag{1}$$

Note that if in this proof we had assumed $c>1$ then $(1)$ would be $0<c^{1/n}-1<\epsilon$, and this proof could have been used for part $(a)$. We assumed, however that $c>0$.

How do we complete this proof? Is it possible with this proof by contradiction? Is there another way?

xoux
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  • I can't follow what you wrote for part (a) at all. – Suzu Hirose Jun 12 '22 at 05:00
  • I think part $(a)$ is incorrect, first of all, because of the use of limits even though we are dealing strictly with natural numbers. However, other than that point (which I think is ultimately correct, though not justified properly I guess), what part of the argument is incorrect? @SuzuHirose? – xoux Jun 12 '22 at 05:02
  • see also https://math.stackexchange.com/questions/3348516/if-a0-prove-that-lim-a1-n-1/3348533#3348533 – AlvinL Jun 12 '22 at 05:15
  • It seems that that solution uses sequences. I am following Spivak chapter by chapter, and sequences comes later in the book in chapter 22. I am looking solutions that rely on basic notions of limits. – xoux Jun 12 '22 at 05:44
  • This is a sequence and the solution proves the limit of the sequence. – AlvinL Jun 12 '22 at 07:54
  • @AlvinL Sure, I get it. I'm just saying that if you assume a reader is going from start to finish through this calculus book he doesn't know what sequences are let alone what the limit of a sequence is. – xoux Jun 12 '22 at 08:02
  • Even the chapter's title is "Limits". Surely basic definitions are provided. – AlvinL Jun 12 '22 at 08:07
  • The solution manual solution for this problem reads: "(This is really a somewhat different kind of "limit at infinity" - which we will study specifically in Chapter 22 - since we are only considering those large $x$ that are natural numbers; actually we will also be able to say that $\lim\limits_{x \to \infty} c^{1/x}=1$, once we have a suitable definition of such powers, in Chapter 18.)" – xoux Jun 12 '22 at 08:16

1 Answers1

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It's good to prove first the case (a) and then the case (b), since (b) follows from (a) immediately.

Proof of (b)

  • If $c>1$ than we are done.
  • If $c=1$ then $c^{1/n}=1\to 1$.
  • If $0<c<1$ then $1/c>1$ and from (a) we get $$\frac 1{c^{1/n}} = \left(\frac 1c\right)^{\frac 1n}\to 1\quad \text{ so }\quad c^{\frac 1n}=\frac{1}{\frac 1{c^{1/n}}}\to \frac 11 = 1.$$
Mateo
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