An Implication of a limit rule

  • Thread starter Rayquesto
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  • #1
Rayquesto
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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)
 

Answers and Replies

  • #2
SammyS
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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 ?
 
  • #3
Rayquesto
318
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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?
 

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