Can i use bernoulis inequality like this?

[Nerpilis]

can I use bernoulis inequality like this for finding this limit?

\lim_{n\rightarrrow\infty}(1+\frac{1}{2n}) \geq \lim (1+2n(\frac{1}{2n})) \geq 2

 

[HallsofIvy]

Well,first of all, you aren't using Bernoulli's inequality.
Bernoulli's inequality says that
(1+x)^r\ge 1+rx
as long as x and r are both larger than -1.

In this case, r= 1 so Bernoulli's inequality doesn't say anything.
In any case, isn't it obvious that, as x->\infty,
\frac{1}{2n}-> 0?
And so
[frac]1+ \frac{1}{2n}\rightarrow 1[/tex].

Or did you mean
lim_{n\rightarrow\infty}\left(1+\frac{2n}\right)^{2n}?
Yes, Bernoulli's inequality applies to this and shows that the limit is greater than 2. In fact, it should be clear that the limit is e which is certainly larger than 2!

 

[Nerpilis]

your right i did have a typo i meant
\lim_{n\rightarrrow\infty}(1+\frac{1}{2n})^{2n} \geq \lim (1+2n(\frac{1}{2n})) \geq 2