# Thread: Limit (Where does f(x) go, when k goes towards infinity?) Answer given, but why?

1. ## Limit (Where does f(x) go, when k goes towards infinity?) Answer given, but why?

Hey math'ers,

I have $\displaystyle I_k= \left( \ln \left( 1+{k}^{-1} \right) \right) ^{k}$, where k goes towards infinity...

I know the answer is supposed to be ln e = 1.

I would think $\displaystyle {k}^{-1}$ goes towards $\displaystyle 0$, which means the expression inside $\displaystyle \ln$ would go towards $\displaystyle 1$, so $\displaystyle \ln$ goes towards $\displaystyle 0$ and the power of $\displaystyle k$ would just speed up the process.

What am I doing wrong?
Any help is appreciated...

Simon DK

2. For the problem you posted, the answer is indeed 0, but I think you meant:

$\displaystyle I_k= \ln \left( \left( 1+{k}^{-1} \right) ^{k} \right)$

Then the expression within ln is the standard limit (or definition) of e, so ln(e)=1.

3. Hello, Simon!

TD! is absolutely correct.

You're expected to know the defintion of $\displaystyle e:$

. . $\displaystyle \lim_{k\to\infty}\left(1 + \frac{1}{k}\right)^k \;=\;e$

I have: .$\displaystyle \lim_{k\to\infty}\left[ \ln \left( 1+{k}^{-1} \right) ^ k\right]$

I know the answer is supposed to be $\displaystyle \ln e = 1$

We have: .$\displaystyle \lim_{k\to\infty}\left[ \ln \left(1+\frac{1}{k}\right) ^ k\right] \;= \;\ln\underbrace{\left[\lim_{k\to\infty} \left( 1+\frac{1}{k}\right)^k\right]}_{\text{this is }e} \;= \;\ln(e) \;=\;1$

4. ## Thanks a whole lot...

Hey there,

I'm sorry, I haven't worked on my math project for a little while as it is pre-exam test season now, so I've not been in time to really look at the answer, but it was a great help - thanks, both of you!

I furthermore notice a few Swedes in the forum questioning about convergence, which I find absolutely perfect. I'm sitting on the side-line looking through answers. ; o ) It must the season of convergence in Scandinavia?

Simon DK