lim ln[1-1/(n+3)] in inf?

[Onixa]

Hi,
What is the result of this limit:

lim ln[1-1/(n+3)] in inf?

is there any equivalent like this:

lim ln[1-1/(n+3)] in inf = lim 1/(n+3) in inf ?

Ans if there is, why?
Tnx

 

[pappus]

Hi,
What is the result of this limit:

lim ln[1-1/(n+3)] in inf?

is there any equivalent like this:

lim ln[1-1/(n+3)] in inf = lim 1/(n+3) in inf ?

Ans if there is, why?
Tnx

Since \(\displaystyle \displaystyle{\frac1{n+3} < \frac1n}\) and \(\displaystyle \displaystyle{\lim_{n \to \infty}\left(\frac1n \right) = 0}\) then

\(\displaystyle \displaystyle{\lim_{n \to \infty}\left(\frac1{n+3} \right) = 0}\).

Use the property of limits:

\(\displaystyle \displaystyle{\lim_{n \to \infty}(a+b) = \lim_{n \to \infty}(a) + \lim_{n \to \infty}(b) }\)

 

[HallsofIvy]

Hi,
What is the result of this limit:

lim ln[1-1/(n+3)] in inf?

is there any equivalent like this:

lim ln[1-1/(n+3)] in inf = lim 1/(n+3) in inf ?

Ans if there is, why?
Tnx

I'm not sure what you mean by "equivalent". Certainly the equation you give is NOT true, because you have dropped both the "ln" and the "1- ".

What is true is that ln (natural logarithm) is continuous for all positive numbers. That means that, as long as 1- 1/(n+3) is positive, \(\displaystyle \lim_{n\to \inf}(1- \frac{1}{n+3})= ln\left(\lim_{n\to\inf}\left(1- \frac{1}{n+3}\right)\right)\)

And that last limit is 1 just as pappus says. So the original limit is ln(1). What is that?