I dont know how to do this limit...
$\displaystyle \lim_{n \rightarrow 0} {\frac{n+1}{2n}}$
Hello, chancey!
The first step (always) is to substitute the limit-value into the functionI 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" $\displaystyle \infty$
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: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
Hello, chancey!
If that's true, you mistyped the original problem.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.
No, it does not exist.Originally Posted by chancey
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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.)