How to calculate the limit of (n^(1/n)-1)* ln n as n goes to infinity

Question

terenzreignz · Answer

$$\lim _{n \rightarrow \infty}[n^{\frac{1}{n} - 1}\ln n]$$
Is this what you meant?

ash2326 · Answer

$$\lim _{n \rightarrow 1}[n^{\frac{1}{n} - 1}\ln n]$$
I think n$\to 1$

anonymous · Answer

No $$\lim_{n \rightarrow \infty} (\sqrt[n]{n} -1) \ln n$$

terenzreignz · Answer

Well, as n goes to infinity, ln n goes to infinity and so does the nth root of n, so..

terenzreignz · Answer

no wait

anonymous · Answer

The answer should be 0, but I don't know how to get there

terenzreignz · Answer

Ok, I can get there, but it require knowledge of L'Hopital's rule, are you familiar with it?

amistre64 · Answer

1/n approaches zero

n^(1/n)  approaches 1

1-1=0

0 lnn = 0 iis my view of it

amistre64 · Answer

Lhop just seems like id be messy :)

terenzreignz · Answer

Lhop is required to prove the n^(1/n) thing, though. It's indeterminate

anonymous · Answer

Yeah, definitely, I know L'hop.

terenzreignz · Answer

Well, as amistre64 said, the limit of n^(1/n) is 1, but can you show it? :)

anonymous · Answer

Yes, I can do that

terenzreignz · Answer

so, from there, you'll get (1-1)ln n
which is just 0 ln n

anonymous · Answer

Yeah, this is where I'm stuck. Getting out of the zero multiplied by inf part

terenzreignz · Answer

It's not 0 times infinity, it's 0 times ln n, which is just 0 :)

anonymous · Answer

but isn't ln n as n goes to infinity, infinity

terenzreignz · Answer

the limit of n as n goes to infinity is also infinity

But I don't think you're about to say that

lim 0n = well, something indeterminate :)

anonymous · Answer

So, would a sequence like this diverge or converge then?!

terenzreignz · Answer

oh it's a sequence?
I thought it was a function >.>
the sequence converges, because the limit exists, in fact, it converges to 0 :)

anonymous · Answer

Yeah, it's a sequence. Sorry, I guess I should've mentioned that part. Sorry, but I still don't get the part about 0 lnn. So, wnevever I get to 0 multiplied by another part that goes to inifinity, I should just treat inf like any other number and so the multiplication would be zero?

terenzreignz · Answer

as long as it's plainly 0 and not some function that just goes to 0 as n approaches infinity

anonymous · Answer

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