I dont know how to do this limit...

$\displaystyle \lim_{n \rightarrow 0} {\frac{n+1}{2n}}$

Printable View

- Jun 7th 2006, 12:25 AMchanceyHelp with limit
I dont know how to do this limit...

$\displaystyle \lim_{n \rightarrow 0} {\frac{n+1}{2n}}$ - Jun 7th 2006, 03:25 AMSoroban
Hello, chancey!

Quote:

I dont know how to do this limit...

$\displaystyle \lim_{n\to0}{\frac{n+1}{2n}}$

. . and see if we get a "reasonable" result.

As $\displaystyle n\to0,\;\frac{n + 1}{2n} \to \frac{1}{0}$ . . . which "equals" $\displaystyle \infty$ - Jun 7th 2006, 03:35 AMchancey
I happen to know that the answer is $\displaystyle \frac{1}{2}$. I remember something ages ago about getting rid of the pronumeral, by dividing by itself?

$\displaystyle \lim_{n \rightarrow 0} {\frac{\frac{n}{n} + \frac{1}{n}}{\frac{2n}{n}}} = \lim_{n \rightarrow 0} {\frac{1 + \frac{1}{n}}{2}} = \frac{1}{2}$

The $\displaystyle \frac{1}{n}$ disapears becuase it is undefined. Thats the method a vaigly remember, but I not sure if i'm right... - Jun 7th 2006, 03:47 AMtopsquarkQuote:

Originally Posted by**chancey**

n = 1/10: $\displaystyle \frac{1+\frac{1}{\frac{1}{10}}}{2}=11/2$

n = 1/100: $\displaystyle \frac{1+\frac{1}{\frac{1}{100}}}{2}=101/2$

n=1/1000: $\displaystyle \frac{1+\frac{1}{\frac{1}{1000}}}{2}=1001/2$

etc. I think you can see that as n approaches 0 the expression blows up.

The trick you are trying to use would work if we were taking the limit as n approaches infinity. In that case 1/n is negligable and can be approximated as 0.

-Dan - Jun 7th 2006, 03:51 AMchanceyQuote:

Originally Posted by**topsquark**

EDIT: When I graph it, and apprach x towards 0, y = 0 ...? - Jun 7th 2006, 04:03 AMSoroban
Hello, chancey!

Quote:

I happen to know that the answer is $\displaystyle \frac{1}{2}$.

It must be: .$\displaystyle \lim_{n\to\infty}\frac{n + 1}{2n}$

. . . . . . . . . . .$\displaystyle \uparrow$

An undefined quantity doesn't simply "disappear".

. . It*stays*in the problem . . . and gives us headaches.

If you don't see that: $\displaystyle \frac{1}{\text{very small number}} \;=\;\text{very large number}$

. . no amount of explaining will change your mind-set. - Jun 7th 2006, 04:08 AMchancey
Ahhhhh Sorry, I wrote it down messy, it is n->inf.

OK, now I know how to do the limit, but is that limit (n->0) even possible? - Jun 7th 2006, 10:55 AMThePerfectHackerQuote:

Originally Posted by**chancey**

---

(However, there are some people that have extended the real number system to include infinity and negative infinity, as well. It is useful when dealing with some limits, like this one here. I personally do not like it because it does not create a field-a very powerful algebraic structure.)