Standard Limits: x^n and x^{1/n}

x^n

sinks to

0

when

|x| < 1

;

x^{1/n}

gathers at

1

when

x > 0

. At

x = -1

it oscillates, and for

|x| > 1

it diverges. Bernoulli's inequality handles every case.

Examples & Exercises

Prove completely by the $\varepsilon$ - $N$ argument that $x^n \to 0$ when $|x| < 1$ . Hint: put $r = 1/|x| > 1$ , write $r = 1 + h$ , and use Bernoulli's inequality $(1+h)^n \ge 1 + nh$ .

r = \frac{1}{\lvert x\rvert} > 1, \quad r = 1 + h, \quad \lvert x\rvert^n \le \frac{1}{1 + nh} \to 0

x = 0.5

n = 20

: the dots sink toward

0

—

0.5^{20} \approx 9.54\times 10^{-7}

0.5^{60} \approx 8.67\times 10^{-19}

Show proof

The case $x = 0$ is trivial, so assume $0 < |x| < 1$ . Put $r = 1/|x|$ ; then $r > 1$ , so $r = 1 + h$ with $h > 0$ .
Bernoulli's inequality $(1+h)^n \ge 1 + nh$ (for $h > -1$ ): equality at $n = 1$ , and if $(1+h)^n \ge 1 + nh$ then $(1+h)^{n+1} = (1+h)^n(1+h) \ge (1+nh)(1+h) = 1 + (n+1)h + nh^2 \ge 1 + (n+1)h$ , so induction gives it. Hence $|x|^n = 1/r^n = 1/(1+h)^n \le 1/(1+nh)$ .
Given $\varepsilon > 0$ , the Archimedean property gives a natural $N > (1/\varepsilon - 1)/h$ . For $n > N$ , $nh > 1/\varepsilon - 1$ , so $1 + nh > 1/\varepsilon$ , i.e. $1/(1+nh) < \varepsilon$ . Hence $|x^n - 0| = |x|^n \le 1/(1+nh) < \varepsilon$ — $x^n \to 0$ .

Prove completely that $x^{1/n} \to 1$ when $x > 1$ . Hint: put $t_n = x^{1/n} - 1 \ge 0$ and apply Bernoulli's inequality to $x = (1+t_n)^n$ .

t_n = x^{1/n} - 1 \ge 0, \qquad x = (1+t_n)^n \ge 1 + nt_n, \qquad 0 \le t_n \le \frac{x-1}{n} \to 0

x = 2

n = 20

2^{1/n}

descends to

1

from above —

2^{1/20} \approx 1.0353

2^{1/60} \approx 1.0116

Show proof

Put $t_n = x^{1/n} - 1$ . Since $x > 1$ , $x^{1/n} > 1$ and $t_n \ge 0$ . By definition $x = (1 + t_n)^n$ .
By Bernoulli's inequality (problem 1), $(1+t_n)^n \ge 1 + nt_n$ , so $x \ge 1 + nt_n$ , i.e. $0 \le t_n \le (x-1)/n$ .
Since $(x-1)/n \to 0$ , the squeeze theorem gives $t_n \to 0$ , i.e. $x^{1/n} = 1 + t_n \to 1$ . (In $\varepsilon$ - $N$ form: for $N > (x-1)/\varepsilon$ , $n > N \Rightarrow |x^{1/n} - 1| = t_n \le (x-1)/n < \varepsilon$ .)

Prove completely that $n^{1/n} \to 1$ . Hint: put $t_n = n^{1/n} - 1$ and keep only the $\binom{n}{2}t_n^2$ term of the binomial expansion.

t_n = n^{1/n} - 1, \qquad n = (1+t_n)^n \ge \binom{n}{2}t_n^2, \qquad 0 \le t_n \le \sqrt{\frac{2}{n-1}} \to 0

x = 2

n = 60

: shows

2^{1/k} \to 1

. The proof below applies the same binomial trick with

x = n

to get

t_n \le \sqrt{2/(n-1)}

(at

n = 60

the bound is

0.1841

, the actual

t_{60} \approx 0.0706

Show proof

Put $t_n = n^{1/n} - 1$ . Since $n \ge 1$ , $t_n \ge 0$ , and by definition $n = (1+t_n)^n$ .
For $n \ge 2$ , the binomial theorem gives $(1+t_n)^n = \sum_{k=0}^{n}\binom{n}{k}t_n^k \ge \binom{n}{2}t_n^2$ (the dropped terms are all nonnegative). So $n \ge \frac{n(n-1)}{2}t_n^2$ .
Dividing by $n > 0$ gives $1 \ge \frac{n-1}{2}t_n^2$ , i.e. $0 \le t_n \le \sqrt{2/(n-1)}$ . Since $\sqrt{2/(n-1)} \to 0$ , the squeeze theorem gives $t_n \to 0$ , i.e. $n^{1/n} = 1 + t_n \to 1$ . Numerical check: at $n = 60$ the bound is $\sqrt{2/59} \approx 0.1841$ and the actual $60^{1/60} - 1 \approx 0.0706$ sits inside it.