Spivak, Ch 5 - Limits Problem 41: For $c>1$, how to show that $c^{1/n} = \sqrt[n]{c}$ approaches $1$ as $n$ > becomes very large?

Question

This is problem 41 from Spivak's Calculus:

(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.

This chapter is the first in the book involving actual Calculus, therefore we only know certain properties of numbers and the definition of a limit in terms of $\epsilon$ and $\delta$.

This problem seems straightforward if the concept of logarithm would be available. I'm not sure if this problem presupposes that this concept is available, however.

The solution manual "solution" says the following:

How do we know that $\sqrt{a^2-\epsilon}$ and $\sqrt{a^2+\epsilon}$ exist!? In Chapter 7 we prove that every positive number has a square root, but the proof of this theorem uses the fact that $f(x)=x^2$ is continuous, which is essentially what we are trying to prove. In fact, the existence of square roots is essentially equivalent to the continuity of $f$ - compare Problem 8-8.

This seems like more of a comment about the problem rather than a solution.

What is a solution to this problem without using logarithms?

EDIT: the solution manual "solution" above is actually from the same numbered problem of an older edition of the book "Calculus". The actual solution is as follows:

If we had $c^{1/n}>1+\epsilon$ then we would have

$$c > (1+\epsilon)^n \geq 1 +n\epsilon$$

or $$n < \frac{c-1}{\epsilon}$$

by the trivial case of Bernoulli's inequality, Problem 2-19, showing that $1<\sqrt[n]{c}<1+\epsilon$ for $n \geq (c-1)/\epsilon$.

Answer 1

Considering comment from user Gary about Bernoulli's Inequality, I came up with the following solution:

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

Since$\lim\limits_{n \to \infty} (1+n\epsilon) = \infty$ we know that $\forall \epsilon>0\ \lim\limits_{n \to \infty} (1+\epsilon)^n = \infty$.

This means that

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

Let $M=c$

Then

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

$$(1+\epsilon)^n>c \implies c^{1/n}-1<\epsilon$$

Since we assume $c>1$ we can also assert that $|c^{1/n}-1|<\epsilon$.

Therefore, what we've shown is

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

Therefore

$$\lim\limits_{n \to \infty} c^{1/n}=1$$

Answer 2

Let $\epsilon >0$ be given.

Assume

$c^{1/n}>1+\epsilon$ for large enough $n$.

Then $c >(1+\epsilon)^n >1+n\epsilon >n\epsilon$;

$r:= c/\epsilon >n$;

Archimedean principle:

Given any real $r>0$ there is an $n_0(\epsilon) \in \mathbb{N}$ s.t.

$n_0 >r$, or $n_0 \epsilon > c$, a contradiction.

Hence for $n\ge n_0:$

$c^{1/n}\le (1+\epsilon);$

and with $1<c^{1/n}$ we get

$1<c^{1/n}\le 1+\epsilon$.