study the continuity of this function

Felafel

1. The problem statement, all variables and given/known data

Study the continuity of the function defined by:
## \lim n \to \infty \frac{n^x-n^{-x}}{n^x+n^{-x}}##

3. The attempt at a solution

I've never seen a limit like this before.
The only thing I have thought of is inserting random values of x to see it the limit exists.
For instance, in this case, for x=0 I'd have ##\frac{\infty^0-\infty^0}{\infty^0+\infty^0}##

which means the function doesn't exist. but every other value of x, it's okay.
Or am i supposed to solve the limit? (btw, how can I solve a limit for n to infinity??)
thank you

jbunniii

By definition, [itex]n^0 = 1[/itex] for any nonzero [itex]n[/itex], so your fraction reduces to
[tex]\frac{n^0 - n^0}{n^0 + n^0} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0[/tex]
What does this imly about the limit
[tex]\lim_{n \rightarrow \infty} \frac{n^0 - n^0}{n^0 + n^0}[/tex]?

Felafel

well, I'd say it means the limit exists and it is = 0 (because I can evaluate it before adding the infinities in the equation).
so the function should continuos on all ##\mathbb{R}##.
Or actually, how should i work for x→∞?
sorry if i have so many doubts about a simple question, but i have never seen this kind of limits before

Dick

It's not that hard. You just have to think through all the cases. Suppose x=1. Try and figure it out without just substituting 'infinity' for n. That's not very informative. Just put x=1.

Felafel

ok.
I see that whatever x i choose (except x=0), I always get a 0 in the nominator and an infinity in the denominator, so appartently f(x)=0. But I'm not sure about that..

jbunniii

Let's take a simple concrete case, say x = 1. Then
[tex]\frac{n^x - n^{-x}}{n^x + n^{-x}} = \frac{n - 1/n}{n + 1/n} = \frac{1 - 1/n^2}{1 + 1/n^2}[/tex]
What is the limit of this expression as [itex]n \rightarrow \infty[/itex]?

Felafel

for x=1 the limit is 1.
and for x=2 as well, because
##\frac{n^2-n^{-2}}{n^2+n^{-2}}##dividing both members by## \frac{1}{n^2} =\frac{1-\frac{1}{n^4}}{1+\frac{1}{n^4}}## and so forth ##\forall x \in \mathbb{R}## except x=0, where f(x)=0.
Then, can i say it is continuos on all |R except between 0 and 1?

Dick

You'd better try some more x values before you state a conclusion. What about x=(1/2) or x=(-1)?

Felafel

Ok.
I've tried several values and found out that if
##x>0 \Rightarrow f(x)=1##
##x=0 \Rightarrow f(x)=0##
##x<0 \Rightarrow \lim f(x)= \frac{\infty}{\infty}## so the function doesn't exist there
Also, it is discontinuos in 0.

Are my assumptions right?

Dick

Can you show how you reached that conclusion for x<0? Try x=(-1). infinity/infinity doesn't necessarily mean the limit is undefined.

Felafel

for x=-1 I get
## \frac{1-\frac{1}{n{^-2}}}{1+ \frac{1}{n^{-2}}}##
##\frac{1-n^2}{1+n^2}= \frac{\infty}{\infty}##
If ##\frac{\infty}{\infty}## doesn't mean it's undefined, when can i say the function is discontinuos?

Dick

Divide numerator and denominator by n^2.

Felafel

right, i didn't notice it, then for x<0 i get f(x)=-1.
what can i say if i get ##\frac{\infty}{\infty}##? and when can i say that the function is not continuos, if that's not enough?

Dick

You can't say anything just from looking at ##\frac{\infty}{\infty}##. A form like that might have a limit and it might not. Ok so f(x)=(-1) for x<0, f(0)=0 and f(x)=1 for x>0. What about continuity?

Felafel

well, as the right limit is different from the left one, I'd say 0 is a discontinuity point (the only one in R).
fortunately, there are no infinity/infinity cases in this function. but if one of these cases happens with a similar limit, should i just leave it? (if algebric manipulation can't help)
thank you very much for your help, anyway :)

Dick

Yes, it's discontinuous only at 0. Leaving a limit as ∞/∞ is about the same thing as saying 'I don't know what the limit is'. If algebra doesn't help you resolve it then there are other tools like l'Hopital.