# Limits of a function containing ln an raised to a power

1. Sep 28, 2008

### dalterego

1. The problem statement, all variables and given/known data
limit x --> infinity, (1+ [(ln2)/x])^x

2. Relevant equations

3. The attempt at a solution

I tried raising the whole thing to e ln [function] so that the x power would get eliminated. What I've got right now is this:

e lim(x->infinity) ln([x+ln2]/2)^x
e lim(x->infinity) xln([x+ln2]/x)
e lim(x->infinity) lnx + ln2
e to the power infinity
= infinity

The limit is not infinity though its 2, I've done it several other ways, but it doesn't get me the right answer anytime.

2. Sep 28, 2008

### foxjwill

I'm not sure how you got to your third step. Since $$\lim_{x\to\infty}x\ln\left(1+\frac{\ln2}{x}\right)$$ is indeterminate, your can rewrite it in the form
$$\lim_{x\to\infty}\frac{\ln\left(1+\frac{\ln2}{x}\right)}{\frac{1}{x}}$$
and then use L'Hopital's rule.

3. Sep 28, 2008

### dalterego

Ok when I use L'Hopital's rule it goes like:
ln([x+ln2]/x)/(1/x)

= (x/[x+ln2])/(-1/x^2) which simplifies to -x^3/(x+ln2)

= -3x^2

which is still infinity

4. Sep 28, 2008

### foxjwill

You forgot about the chain rule.

5. Sep 28, 2008

### HallsofIvy

Staff Emeritus
Do you think that having a log in this is a problem? Would it be any easier if it were just $\lim_{x\rightarrow\infty} (1+ a/x)^x ? If so, ln(2) is just a number, let a= ln(2)! I seem to recall (check me on this!) that $$\lim{x\rightarrow\infty} (1+ 1/x)^x= e$ But that has no "ln(2)" or even "a" in it does it. No matter. If the problem is [tex]\lim{x\rightarrow\infty}(1+ a/x)^x$$
let y= x/a. Then a/x= 1/y and x= ay
$$(1+ a/x)^x= (1+ y)^{ay}$$
and
$$\lim_{x\rightarrow\infty}(1+ a/x)^x= \lim{y\rightarrow\infty(1+ 1/y)^{ay}$$
$$= \left(\lim_{y\rightarrow\infty (1+ 1/y)^y\right)^a= e^a$$

If a= ln(2) then the limit is eln(2). What is that?