Question

What is the limit as n approaches infinity of (3sqrt(n))^(1/2n)
Main reason this frustrates me is cause I know a bunch of easy ways to do this I'm not allowed to use. I cannot use L'hopitals or even ln(f(x)). I have the epsilon definition of a limit and the basic addition/subtraction/multiplication/division limit laws

Answer 1

\[\lim_{n \rightarrow \infty} (3\sqrt{n})^{1/2n}\] is how it appears in my book

Answer 2

Is it \(3\sqrt n\) or \(\sqrt[3]{n}\) ?

Also, is it \(\dfrac{1}{2n}\) or \(\dfrac{1}{2}n\) ?

Answer 3

I know its \[3\sqrt{n}\]
I believe its \[\frac{ 1 }{ 2n }\] as otherwise the limit is divergent (+infinity) and the question is phrased "Determine the following limits". Forget to mention I do have squeeze theorem

Answer 4

Those instructions don't seem to suggest that the limit must exist or be finite

Answer 5

\[\lim_{n \rightarrow \infty} (3\sqrt{n})^{1/2n}\]
is how it appears in my book.

Answer 6

I keep getting it to the point where I just need to prove that
\[\lim_{n \rightarrow \infty} (9n)^{(1/n)}=1\] but I can't get there.without using one of the methods I'm not allowed to.

Answer 7

So you have to both find the limit and prove it?

Answer 8

Well I have to prove that the limit I found is correct, but that is usually proven in the work finding the limit. I may not be understanding you here.

Answer 9

Allow me to clarify: evaluating the limit is a process completely distinct from proving a limit. Evaluation is done with algebraic manipulation until you have something that can be evaluated directly (provided the expression is continuous), while a proof would involve applying the \(\epsilon-\delta\) definition to establish that \(\lim_{n\to\infty}f(n)=1\).

Answer 10

Ah, I can't pull 1 out of a straw hat unfortunately so we need to evaluate it. If at some point it becomes a rather simple limit where it is foreseeable without using the before mentioned out-lawed methods I may use the epsilon definition to confirm it.

Answer 11

I hope that makes sense.

Answer 12

You're not given a lot of freedom here... L'Hopital's would have been of great use. The algebraic properties of limits won't be much help. The squeeze theorem would probably be the most useful way to approach this, but it'll be tricky to come up with bounding sequences.

Answer 13

Aye, I want to use the sequence (1) as the lower bound but an upper bound sequence that goes to 1....is rather difficult.

Answer 14

I suppose there's nothing stopping you from using \(a_n=1-\dfrac{1}{n}\) and \(b_n=1+\dfrac{1}{n}\) (since we know that the limit must be 1.

Answer 15

Or \(a_n=1\) and \(b_n=1+\dfrac{1}{n}\).

Answer 16

alright, give me a sec to try and work on a proof that \[b _{n} \ge x _{n}\]
for sufficiently large n

Answer 17

You would then only need to show that for all \(n\), or at least all \(n\) after some large enough \(N\), the following holds.
\[\large(3\sqrt n)^{\frac{1}{2n}}\le1+\frac{1}{n}\]

Answer 18

I might suggest a proof by induction to establish this, but that may be easier said than done.

Answer 19

definitely easier said then done, might try it on the equation
\[(9n)^{1/n}\] as I do have a theorem that if \[\lim_{n \rightarrow \infty} x _n = c \rightarrow \lim_{n \rightarrow \infty} \sqrt x _n = \sqrt c\]
there are some other conditions but this sequence satisfies them