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Graphically, I see that $\lim_{n->\infty}3^{1/n}$ approaches $1$. However, how to show $\lim_{n->\infty}3^{1/n} = 1$ step by step?

Bernard
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user214969
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7 Answers7

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For any number $a>0$, $a^{1/n}$ tends to $1$ when $n$ tends to infinity since $$\log a^{1/n}=\frac1n\,\log a\to 0\enspace\text{when }\enspace n\to\infty.$$

Bernard
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  • Nice approach Bernard! – Mehrdad Zandigohar Feb 13 '18 at 20:07
  • nice indeed Bernard +1.. – user577215664 Feb 13 '18 at 20:24
  • Just be careful - that assumes that $\log(x)$ is continuous at $x=1.$ – Thomas Andrews Feb 13 '18 at 20:38
  • @ThomasAndrews: Of course. But you need the log (and its continuity) to prove the existence of $n$-th roots. – Bernard Feb 13 '18 at 20:41
  • @Bernard No, you don't. You just need the LUB property of the reals. $3^{1/n}=\sup{x\in\mathbb R^{+}\mid x^n\leq 3}.$ You can prove that $x$ is non-empty, and has an upper bound, so $3^{1/n}$ is real. Then you can prove that if $x^n<3$ then there is a $y>x$ such that $y^n<3,$ hence $(3^{1/n})^n=3.$ These are all definable without logarithms. – Thomas Andrews Feb 13 '18 at 20:45
  • The only continuity you need is that $(1+x)^n$ is continous at $x=0.$ – Thomas Andrews Feb 13 '18 at 20:50
  • You can, of course, with the l.u.b. property. But it becomes very complex, especially when you want to prove differentiability, and anyway the log is used for non-rational exponents. That's why usually one uses the log to establish the existence of $n$-th roots. – Bernard Feb 13 '18 at 20:57
  • Sure, but when you get to that point, you can prove the LUB property is the same as the $a^x=e^{x\log a}$ definition absolutely trivially. In reality, most students simply "know" that $3^{1/n}$ exists, if they've encountered the notation rather than $\sqrt[n]{3}$. One doesn't actually need the definition, only that there is a unique positive $n$th root. – Thomas Andrews Feb 13 '18 at 20:59
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Note that

$$\lim_{n->\infty}\frac1n = 0 \quad \quad 3^0=1$$

thus by algebraic rule and continuity

$$\lim_{n->\infty}3^{1/n}=3^{\lim_{n->\infty}{1/n}}$$

we have

$$3^{1/n}\to 3^0=1$$

user
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You only need two elementary results: The binomial theorem and the squeeze theorem.

The binomial theorem implies that if $x>0$ then $(1+x)^n\geq 1+nx$.

Letting $x_n=3^{1/n}-1$, then $$(1+x_n)^{n}=3.$$ Since $x_n>0$, we have, by the above result, that $$3=(1+x_n)^n \geq 1+nx_n.$$

So $0\leq x_n\leq\frac{2}{n}$. Hence $x_n\to 0$ by the squeeze theorem, and and hence $3^{1/n}=x_n+1\to 1.$

This works for any sequence $a^{1/n}$ with $a>1$ since $x_n=a^{1/n}-1$ gives us that $0<x_n\leq \frac{a-1}{n}.$

To show it for $0<a<1$, we just need that $f(x)=\frac{1}{x}$ is continuous at $x=1.$ Assuming that, we use that $$ a^{1/n}=f\left(\left(a^{-1}\right)^{1/n}\right)$$ And if $0<a<1$, then $1<a^{-1}$ and by our previous result: $$\left(a^{-1}\right)^{1/n}\to 1.$$ So $a^{1/n}\to f(1)=1.$

Thomas Andrews
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  • And Bernoulli's Inequality, which is all you really used here, is even more basic than the binomial theorem. Both can easily be proven with induction, but the induction proof for BI is almost trivial. – Mark Viola Feb 13 '18 at 20:43
  • True, but most students don't learn Bernoulli when they are learning calculus, so I wanted to avoid calling out to it. But I had forgotten how simple the proof of Bernoulli actually is. – Thomas Andrews Feb 13 '18 at 20:57
  • Your answer is a very good one. (+1) I only wanted to point out the seemingly lost (as you implied) use of BI. – Mark Viola Feb 13 '18 at 21:09
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HINT:

Using $e^x\le \frac1{1-x}$ for $x<1$, we have for $n\ge 2$

$$1< 3^{1/n}< \frac{1}{1-\frac1n \log(3)}$$

Now apply the squeeze theorem.

Mark Viola
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Binomial theorem:

For $x\ge 0:$

$(1+x)^n \ge 1 + nx + n(n-1)\dfrac{x^2}{2!} \ge $

$n^2\dfrac{x^2}{4}$ for $n\ge 2.$

Set $x=\dfrac{2√3}{n}:$

$(1+\dfrac{2√3}{n})^n \ge n^2\dfrac{(4)(3)}{4n^2}= 3;$

$ (1+\dfrac{2√3}{n})^{n} \ge 3$ , or

$1 +\dfrac{2√3}{n} \ge 3^{1/n} .$

With lower bound $1$:

$1\lt 3^{1/n} \le 1+ \dfrac{2√3}{n}.$

Limit $n \rightarrow \infty$ is?

Peter Szilas
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As an alternative, with a more advanced approach by the ratio-root criteria

$$a_n = \sqrt[n]{b_n} \quad b_n=3$$

$$\frac{b_{n+1}}{b_n} \rightarrow L\implies a_n=b_n^{\frac{1}{n}} \rightarrow L$$

thus since

$$\frac{b_{n+1}}{b_n}=\frac33=1 \implies a_n = \sqrt[n]{3}\to 1$$

user
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Bernoulli's Inequality, which, for integer exponents, can be proven using a simple inductive argument, says $$ \left(1+\frac2n\right)^n\ge3\ge1 $$ Taking roots, we get $$ 1\le3^{1/n}\le1+\frac2n $$ Then, the Squeeze Theorem ensures that $$ \lim_{n\to\infty}3^{1/n}=1 $$

robjohn
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