1. ## limit question

what is the limit of $(ln^2 k)^{\frac{1}{k}}$
when k goes to infinity
?

2. ## Re: limit question

Originally Posted by transgalactic
what is the limit of $(ln^2 k)^{\frac{1}{k}}$
when k goes to infinity
?
\displaystyle \begin{align*}\lim_{k \to \infty}\left(\ln^2{k}\right)^{\frac{1}{k}} &= \lim_{k \to \infty}e^{\ln{\left[\left(\ln^2{k}\right)^{\frac{1}{k}}\right]}} \\ &= e^{\lim_{k \to \infty}\ln{\left[\left(\ln^2{k}\right)^{\frac{1}{k}}\right]}} \\ &= e^{\lim_{k \to \infty}\left[\frac{\ln{\left(\ln^2{k}\right)}}{k}\right]}\end{align*}

Now you can apply L'Hospital's Rule.

3. ## Re: limit question

Originally Posted by transgalactic
what is the limit of $(ln^2 k)^{\frac{1}{k}}$ when k goes to infinity?
This can also be done by direct comparison.

If $k\ge 3$ then we get $\ln(k)\le 2\sqrt{k}$.

Thus $1 \leqslant \sqrt[k]{{\ln ^2 (k)}} \leqslant \sqrt[k]{{4k}}$

4. ## Re: limit question

Originally Posted by transgalactic
what is the limit of $(ln^2 k)^{\frac{1}{k}}$
when k goes to infinity
?
Could you please clarify your notation, does $\log^2k$ denote $(\log(k))^2$ or $\log(\log(k))$ ?

Not that it makes a jot of difference to the answer!

CB

5. ## Re: limit question

Originally Posted by CaptainBlack
Could you please clarify your notation, does $\log^2k$ denote $(\log(k))^2$ or $\log(\log(k))$ ?

Not that it makes a jot of difference to the answer!

CB
I expect it's being used in the same context as $\displaystyle \sin^2{x}$ for $\displaystyle (\sin{x})^2$...

6. ## Re: limit question

Originally Posted by Prove It
I expect it's being used in the same context as $\displaystyle \sin^2{x}$ for $\displaystyle (\sin{x})^2$...
So do I, but I object to the abuse of notation in this case because the iterated logarithm is a relativly common function.

(as it happens I also object to the systematic omission of parentheses around the arguments of functions in elementary mathematics)

CB