Question

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.

Answer 1

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

Answer 2

and yes as x goes off to infinity, so does y, but not very fast

Answer 3

infinity therefore gets bigger faster than ln(infinty) does

Answer 4

so you end up with a smaller number over a bigger number which goes off to zero..

Answer 5

hehe i think...

Answer 6

Thx. Thinking in graph method is more easily comprehended sometimes. :)

Answer 7

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

Answer 8

Thanks. :)

Answer 9

np

Answer 10

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

Answer 11

I don't think I follow why the limit can just jump up on the e. :(

Answer 12

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

Answer 13

strike the "it is okay to do this because" at the bottom there...

Answer 14

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
\[\lim_{x\to a}e^{f(x)}=e^{\lim_{x\to a}f(x)}\]

Answer 15

what turning test said

Answer 16

well, except for the "e is a constant" part. limit goes up because
\[\exp\] is a continuous function

Answer 17

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 !!

Answer 18

Thanks for correcting my reasoning @satellite73 , I shouldn't have said that since I wasn't sure.

Answer 19

Thanks you guys. You're both so helpful. :)

I think I remember the log and ln rules more. Guess I should review some more.