# An Implication of a limit rule

Rayquesto

## Homework Statement

If lim (x->∞) [ln(x^(1/x))]=0 and lim (x->∞) x^(1/x)=1, then does this

=>

lim (x->∞) [ln(x^(1/x))]= ln(lim(x->∞) [(x^(1/x))]) = ln(1)??

## Homework Equations

lim (x->∞) [ln(x^(1/x))]=0 and lim (x->∞) x^(1/x)=1

## The Attempt at a Solution

lim (x->∞) [ln(x^(1/x))]= ln(lim(x->∞) [(x^(1/x))] = ln(1)

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## Homework Statement

If lim (x->∞) [ln(x^(1/x))]=0 and lim (x->∞) x^(1/x)=1, then does this

=>

lim (x->∞) [ln(x^(1/x))]= ln(lim(x->∞) [(x^(1/x))]) = ln(1)??

## Homework Equations

lim (x->∞) [ln(x^(1/x))]=0 and lim (x->∞) x^(1/x)=1

## The Attempt at a Solution

lim (x->∞) [ln(x^(1/x))]= ln(lim(x->∞) [(x^(1/x))] = ln(1)
Just what is the question?

You do realize that ln(1) = 0, don't you ?

Rayquesto
Yes, but the question refers to the limit rules. This is what I am asking:

lim (x->∞) [ln(x^(1/x))]= ln(lim(x->∞) [(x^(1/x))]) = ln(1)??

To put it into words: Can the limit as x approaches infinity of [ln(x^(1/x))] be equal to the natural log of the limit as x approaches infinity of [(x^(1/x))] since lim (x->∞) [ln(x^(1/x))]=0 and lim (x->∞) x^(1/x)=1?