mysesshou
  • mysesshou
Find the limit of nth sqrt(n) So, I was just wondering which method is best and if there is any better way of doing this. Thanks guys.
Mathematics
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SOLVED
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katieb
  • katieb
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eyust707
  • eyust707
okay so im not sure of the techincal way of doing the first one but intuitively if we look at the graph of ln (x) it looks like this
eyust707
  • eyust707
|dw:1332531021653:dw|
eyust707
  • eyust707
and yes as x goes off to infinity, so does y, but not very fast

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eyust707
  • eyust707
infinity therefore gets bigger faster than ln(infinty) does
eyust707
  • eyust707
so you end up with a smaller number over a bigger number which goes off to zero..
eyust707
  • eyust707
hehe i think...
mysesshou
  • mysesshou
Thx. Thinking in graph method is more easily comprehended sometimes. :)
eyust707
  • eyust707
looks like your better than i at these but once again for the second one i would just use intutition... as you said: the root n of n = n^(1/n) , as n goes to infinity th root becomes 0. anything to the zero power is 1
mysesshou
  • mysesshou
Thanks. :)
eyust707
  • eyust707
np
TuringTest
  • TuringTest
the way these should be done are on the attachments at the top of the thread the first problem is done using l'Hospitals rule the second is done by rewriting it as\[\huge \lim_{n\to\infty}\sqrt[n]n=e^{\lim_{n\to\infty}\ln \sqrt[n]n}\]we can then use log properties, then l'hospital
mysesshou
  • mysesshou
I don't think I follow why the limit can just jump up on the e. :(
TuringTest
  • TuringTest
because we have that\[\huge a^{\log_a x}=x\]the trick is to go\[\huge \lim_{n\to a}f(x)=\lim_{n\to a}e^{\ln f(x)}\]since e is a constant we can just move the limit to the exponent\[\huge=e^{\lim_{n\to a}\ln f(x)}\]then hopefully you can evaluate the exponent it is okay to do this because
TuringTest
  • TuringTest
strike the "it is okay to do this because" at the bottom there...
anonymous
  • anonymous
the "e thing" is because the precise definition of \[b^x\] is \[e^{x\ln(b)}\] and you can take the limit up into the exponent because \[e^x\] is a continuous function, meaning \[\lim_{x\to a}e^{f(x)}=e^{\lim_{x\to a}f(x)}\]
anonymous
  • anonymous
what turning test said
anonymous
  • anonymous
well, except for the "e is a constant" part. limit goes up because \[\exp\] is a continuous function
mysesshou
  • mysesshou
Ah, right. I'm just slow and blind. I missed the ln in the exponent. I should have remembered the ln and e rules better. The last time I used them I had to look them up. Thanks so much. You're the best !!
TuringTest
  • TuringTest
Thanks for correcting my reasoning @satellite73 , I shouldn't have said that since I wasn't sure.
mysesshou
  • mysesshou
Thanks you guys. You're both so helpful. :) I think I remember the log and ln rules more. Guess I should review some more.

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