# Prove the remarkable limit of the function a^1/n, a>0, lim a^1/n=1 if n approaches to + infinite

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### 2 Answers

We have to find the value of lim n-->+inf. [ a^(1/n)]

We know that as n tends to infinity 1/n tends to 0.

lim n-->+inf. [ a^(1/n)]

=> lim (1/n)-->0 [ a^(1/n)]

Any number raised to the power 0 gives 1. Substituting infinity for n we get:

a^(0) = 1

**Therefore the value of lim n-->+inf. [ a^(1/n)] = 1.**

To prove this remarkable limit, we'll prove that the limit of ratio of 2 consecutive terms of the sequence (an) exists and it's value is 1.

lim a n+1/a n = lim a^[1/(n+1)]/a^(1/n)

We'll use the quotient property of exponentials that have matching bases:

lim a^[1/(n+1)]/a^(1/n) = lim a^[1/(n+1) - (1/n)]

lim a^[1/(n+1) - (1/n)] = lim a^[-1/n(n+1)]

We'll use the property of negative power:

lim a^[-1/n(n+1)] = lim 1/a^[1/n(n+1)]

lim 1/a^[1/n(n+1)] = 1/a^lim [1/n(n+1)]

lim [1/n(n+1)] = 1/+infinite = 0

1/a^lim [1/n(n+1)] = 1/a^0 = 1/1 = 1

The limit of the ratio a n+1/ a n exists and it is 1.

**Therefore, the remarkable limit, if n approaches +infinite, is: lim a^(1/n) = 1.**