Limits of a function containing ln an raised to a power

In summary: Yes, you are correct. The limit can be rewritten as (1+1/x)^x, which is equal to e^ln(2), which simplifies to 2. This makes sense since as x approaches infinity, the ln(2)/x term becomes smaller and smaller, essentially approaching 0, and (1+0)^x is just 1. So the limit is 2. In summary, the limit of (1+(ln2)/x)^x as x approaches infinity is equal to 2.
  • #1
dalterego
14
0

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.
 
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  • #2
I'm not sure how you got to your third step. Since [tex]\lim_{x\to\infty}x\ln\left(1+\frac{\ln2}{x}\right)[/tex] is indeterminate, your can rewrite it in the form
[tex]\lim_{x\to\infty}\frac{\ln\left(1+\frac{\ln2}{x}\right)}{\frac{1}{x}}[/tex]
and then use L'Hopital's rule.
 
  • #3
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
You forgot about the chain rule.
 
  • #5
Do you think that having a log in this is a problem? Would it be any easier if it were just [itex]\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
[tex]\lim{x\rightarrow\infty} (1+ 1/x)^x= e[/itex]

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[/tex]
let y= x/a. Then a/x= 1/y and x= ay
[tex](1+ a/x)^x= (1+ y)^{ay}[/tex]
and
[tex]\lim_{x\rightarrow\infty}(1+ a/x)^x= \lim{y\rightarrow\infty(1+ 1/y)^{ay}[/tex]
[tex]= \left(\lim_{y\rightarrow\infty (1+ 1/y)^y\right)^a= e^a[/tex]

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

1. What is the definition of a limit of a function containing ln an raised to a power?

The limit of a function containing ln an raised to a power is the value that the function approaches as the input (x) approaches a certain value. In other words, it is the value that the function "approaches" but may not necessarily equal at that specific input value.

2. How do you find the limit of a function containing ln an raised to a power?

To find the limit of a function containing ln an raised to a power, you can use the following steps:

  1. Substitute the input value (x) into the function.
  2. Simplify the resulting expression as much as possible.
  3. Check for any values that would make the function undefined (such as dividing by zero).
  4. Finally, take the limit of the simplified expression as x approaches the input value.

3. Can the limit of a function containing ln an raised to a power be undefined?

Yes, the limit of a function containing ln an raised to a power can be undefined. This typically occurs when the function has a vertical asymptote, meaning there is a value that the function approaches but never reaches.

4. Can the limit of a function containing ln an raised to a power be negative?

Yes, the limit of a function containing ln an raised to a power can be negative. The limit is simply the value that the function approaches and does not necessarily have to be positive or negative.

5. How does the limit of a function containing ln an raised to a power relate to the derivative of the function?

The limit of a function containing ln an raised to a power can provide information about the derivative of the function. Specifically, if the limit exists, it can be used to find the derivative of the function at that specific input value by using the definition of the derivative.

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