Limits of a function containing ln an raised to a power

[dalterego]

Homework Statement

limit x --> infinity, (1+ [(ln2)/x])^x

Homework Equations

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.

 

[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.

 

[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

 

[foxjwill]

You forgot about the chain rule.

 

[HallsofIvy]

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)!<br /> I seem to recall (check me on this!) that<br /> \lim{x\rightarrow\infty} (1+ 1/x)^x= e<br /> <br /> But that has no &quot;ln(2)&quot; or even &quot;a&quot; in it does it. No matter. If the problem is <br /> \lim{x\rightarrow\infty}(1+ a/x)^x<br /> let y= x/a. Then a/x= 1/y and x= ay<br /> (1+ a/x)^x= (1+ y)^{ay}<br /> and <br /> \lim_{x\rightarrow\infty}(1+ a/x)^x= \lim{y\rightarrow\infty(1+ 1/y)^{ay}<br /> = \left(\lim_{y\rightarrow\infty (1+ 1/y)^y\right)^a= e^a<br /> <br /> If a= ln(2) then the limit is e^ln(2). What is that?