# Help with limit

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• Jun 7th 2006, 12:25 AM
chancey
Help with limit
I dont know how to do this limit...

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

Quote:

I dont know how to do this limit...

$\displaystyle \lim_{n\to0}{\frac{n+1}{2n}}$
The first step (always) is to substitute the limit-value into the function
. . and see if we get a "reasonable" result.

As $\displaystyle n\to0,\;\frac{n + 1}{2n} \to \frac{1}{0}$ . . . which "equals" $\displaystyle \infty$
• Jun 7th 2006, 03:35 AM
chancey
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 AM
topsquark
Quote:

Originally Posted by chancey
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...

The 1/n canNOT simply disappear. Consider:
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 AM
chancey
Quote:

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 ...?
• Jun 7th 2006, 04:03 AM
Soroban
Hello, chancey!

Quote:

I happen to know that the answer is $\displaystyle \frac{1}{2}$.
If that's true, you mistyped the original problem.

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 AM
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?
• Jun 7th 2006, 10:55 AM
ThePerfectHacker
Quote:

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.

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(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.)