Limit of the n-th Root as n Tends to Infinity
The limit of the n-th root as n tends to infinity depends on the value of the radicand.
If the radicand (a) is greater than or equal to 1, the limit as n→∞ is equal to 1. $$ \lim_{n \rightarrow ∞} \sqrt[n]{a} = 1 $$
The n-th root of a number can equivalently be written in exponential form
$$ \sqrt[n]{a} = a^{\frac{1}{n}} $$
As n tends to infinity, the exponent \(\frac{1}{n}\) tends to zero, so the expression approaches \(a^0\), which is equal to 1 for any nonzero real number.
$$ \lim_{n \rightarrow ∞} a^{\frac{1}{n}} = 1 $$
Proof
If a≥1, then the n-th root is also greater than or equal to 1.
$$ a \ge 1 \Rightarrow \sqrt[n]{a} \ge 1 $$
Hence
$$ \sqrt[n]{a} - 1 \ge 0 $$
Define the sequence \( b_n \) by
$$ b_n = \sqrt[n]{a} - 1 \ge 0 $$
Then
$$ b_n + 1 = \sqrt[n]{a} $$
Raising both sides to the power n gives
$$ (b_n + 1)^n = (\sqrt[n]{a})^n $$
$$ (b_n + 1)^n = a $$
By Bernoulli's inequality
$$ (b_n + 1)^n \ge 1 + n b_n $$
$$ a \ge 1 + n b_n $$
$$ \frac{a - 1}{n} \ge b_n $$
Thus \( 0 \le b_n \le \frac{a - 1}{n} \)
$$ 0 \le b_n \le \frac{a - 1}{n} $$
Taking limits
$$ \lim_{n \rightarrow ∞} 0 \le \lim_{n \rightarrow ∞} b_n \le \lim_{n \rightarrow ∞} \frac{a - 1}{n} $$
$$ 0 \le \lim_{n \rightarrow ∞} b_n \le 0 $$
By the squeeze theorem
$$ \lim_{n \rightarrow ∞} b_n = 0 $$
It follows that
$$ \lim_{n \rightarrow ∞} \sqrt[n]{a} = 1 $$
If the radicand (a) is raised to a power m, the limit as n→∞ remains equal to 1. $$ \lim_{n \rightarrow ∞} \sqrt[n]{a^m} = 1 $$
Proof
Setting m = 1/2, the radicand becomes √a
$$ b_n = \sqrt[n]{a^{1/2}} - 1 \ge 0 $$
$$ b_n = \sqrt[n]{\sqrt{a}} - 1 \ge 0 $$
Then
$$ b_n + 1 = \sqrt[n]{\sqrt{a}} $$
Raising both sides to the power n gives
$$ (b_n + 1)^n = (\sqrt[n]{\sqrt{a}})^n $$
$$ (b_n + 1)^n = \sqrt{a} $$
By Bernoulli's inequality
$$ (b_n + 1)^n \ge 1 + n b_n $$
$$ \sqrt{a} \ge 1 + n b_n $$
$$ \frac{\sqrt{a} - 1}{n} \ge b_n $$
Thus \( 0 \le b_n \le \frac{\sqrt{a} - 1}{n} \)
$$ 0 \le b_n \le \frac{\sqrt{a} - 1}{n} $$
Taking limits
$$ \lim_{n \rightarrow ∞} 0 \le \lim_{n \rightarrow ∞} b_n \le \lim_{n \rightarrow ∞} \frac{\sqrt{a} - 1}{n} $$
$$ 0 \le \lim_{n \rightarrow ∞} b_n \le 0 $$
By the squeeze theorem
$$ \lim_{n \rightarrow ∞} b_n = 0 $$
Therefore
$$ \lim_{n \rightarrow ∞} \sqrt[n]{\sqrt{a}} = 1 $$
And so on
