Compute the limit of (1+xⁿ)^1/n and (aⁿ + b^n1/n

by RoRi

March 21, 2016

Consider the limit
$\lim_{n \to \infty} (1+x^n)^{\frac{1}{n}}.$

For $0 < x < 1$ prove that this limit exists and compute the limit.
For positive real numbers $a$ and $b$ , Consider the limit
$\lim_{n \to \infty} (a^n + b^n)^{\frac{1}{n}}.$

Compute this limit.

Proof. We use the squeeze theorem to prove existence and compute the value of the limit. Since $0<x<1$ we have
$(1+0^n)^{\frac{1}{n}} < (1+x^n)^{\frac{1}{n}} < (1+1^n)^{\frac{1}{n}}$

for all positive integers $n$ . Then we have

$\begin{align*} \lim_{n \to \infty} (1+0^n)^{\frac{1}{n}} &= \lim_{n \to \infty} 1^{\frac{1}{n}} = 1 \\[9pt] \lim_{n \to \infty} (1+1^n)^{\frac{1}{n}} &= \lim_{n \to \infty} 2^{\frac{1}{n}} = 1. \end{align*}$

(We know the second limit from this previous exercise (Section 10.4, Exercise #9).) Therefore, by the squeeze theorem we have

$\lim_{n \to \infty} \left( 1 + x^n \right)^{\frac{1}{n}} = 1. \qquad \blacksquare$
If $a > b$ then we have $0 < \frac{a}{b} < 1$ and so, using part (a),
$\begin{align*} \lim_{n \to \infty} \left( a^n + b^n \right)^{\frac{1}{n}} &= \lim_{n \to \infty} a \cdot \left( 1 + \left( \frac{b}{a} \right)^n \right)^{\frac{1}{n}} \\[9pt] &= a. \end{align*}$

On the other hand if $b > a$ then we have $0 < \frac{a}{b} < 1$ and so by part (a) again,

$\begin{align*} \lim_{n \to \infty} \left( a^n + b^n \right)^{\frac{1}{n}} &= \lim_{n \to \infty} b \cdot \left( 1 + \left(\frac{a}{b}\right)^n \right)^{\frac{1}{n}} \\[9pt] &= b. \end{align*}$

If $a = b$ then

$\begin{align*} \lim_{n \to \infty} (a^n + b^n)^{\frac{1}{n}} &= a \cdot \lim_{n \to \infty} \left( 1 + \left( \frac{b}{a} \right)^n \right)^{\frac{1}{n}} \\[9pt] &= a \cdot \lim_{n \to \infty} 2^{\frac{1}{n}} \\[9pt] &= a = b. \end{align*}$

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