Prove:
Limit as n --> infinity
n/(n^2+3n+1) -->0
I just don't know how to start here. What algebraic property can I use to get started?
Will this work?
|n/(n^2 +3n + 1)| < | n/(n^2 +3n)| < | n/(n^2)| = 1/n?
Thanks.
Prove:
Limit as n --> infinity
n/(n^2+3n+1) -->0
I just don't know how to start here. What algebraic property can I use to get started?
Will this work?
|n/(n^2 +3n + 1)| < | n/(n^2 +3n)| < | n/(n^2)| = 1/n?
Thanks.
This integral is of the indeterminate form
$\displaystyle \frac{\infty}{\infty}$
so L'Hospital's Rule can be used.
$\displaystyle \lim_{n \to \infty}\frac{n}{n^2 + 3n + 1} = \lim_{n \to \infty}\frac{\frac{d}{dx}(n)}{\frac{d}{dx}(n^2 + 3n + 1)}$
$\displaystyle = \lim_{x \to \infty}\frac{1}{2n + 3}$
$\displaystyle \to \frac{1}{\infty}$
$\displaystyle \to 0$.
I assume from your title, and the fact that this thread is posted in a pretty advanced sub-forum, that you wish to formally prove the statement
$\displaystyle \lim_{n\to\infty}\frac{n}{n^2+3n+1}$.
To do this, the epsilon-delta definition is in order.
The statement $\displaystyle \lim_{x\to\infty}f(x)=L$ means that for each $\displaystyle \epsilon>0$, there exists a [MAth]M>0[/tex] such that $\displaystyle |f(x)-L|<\epsilon$ whenever $\displaystyle x>M$.
In your case, it is not difficult to see that for any $\displaystyle \epsilon>0$ a number $\displaystyle M$ can be found such that
$\displaystyle |f(n)-0|<\epsilon$ whenever $\displaystyle n>M$
This implies that the limit is ,in fact, 0.
For the proof of $\displaystyle n\to{-\infty}$, just take values of $\displaystyle n$ which are less than some number $\displaystyle N<0$.
Yes, I do and thanks. I was primarily looking for a way to get started, and I can fill in all the formalities. This is my first analysis class, so all responses are extremely useful, and somewhat ease the frustration of taking such a tedious class.
Thanks everyone!