Help with limit

**chancey** · June 7th 2006, 12:25 AM

I dont know how to do this limit...

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

**Soroban** · June 7th 2006, 03:25 AM

Hello, chancey!

I dont know how to do this limit...

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

The first step (always) is to substitute the limit-value into the function
. . and see if we get a "reasonable" result.

As $n\to0,\;\frac{n + 1}{2n} \to \frac{1}{0}$ . . . which "equals" $\infty$

**chancey** · June 7th 2006, 03:35 AM

I happen to know that the answer is $\frac{1}{2}$ . I remember something ages ago about getting rid of the pronumeral, by dividing by itself?

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

The $\frac{1}{n}$ disapears becuase it is undefined. Thats the method a vaigly remember, but I not sure if i'm right...

**topsquark** · June 7th 2006, 03:47 AM

Originally Posted by chancey

I happen to know that the answer is $\frac{1}{2}$ . I remember something ages ago about getting rid of the pronumeral, by dividing by itself?

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

The $\frac{1}{n}$ disapears becuase it is undefined. Thats the method a vaigly remember, but I not sure if i'm right...

The 1/n canNOT simply disappear. Consider:
n = 1/10: $\frac{1+\frac{1}{\frac{1}{10}}}{2}=11/2$

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

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

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

**chancey** · June 7th 2006, 03:51 AM

Originally Posted by topsquark

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.

But im working out as n approches 0, not infinity?

EDIT: When I graph it, and apprach x towards 0, y = 0 ...?

**Soroban** · June 7th 2006, 04:03 AM

Hello, chancey!

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

If that's true, you mistyped the original problem.

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

An undefined quantity doesn't simply "disappear".
. . It stays in the problem . . . and gives us headaches.

If you don't see that: $\frac{1}{\text{very small number}} \;=\;\text{very large number}$
. . no amount of explaining will change your mind-set.

**chancey** · June 7th 2006, 04:08 AM

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?

**ThePerfectHacker** · June 7th 2006, 10:55 AM

Originally Posted by chancey

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?

No, it does not exist.

---
(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.)

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