Study the continuity of this function

[Felafel]

Homework Statement

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

 

[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]

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

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]

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..

Let's take a simple concrete case, say x = 1. Then
\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}
What is the limit of this expression as n \rightarrow \infty?

 

[Felafel]

Let's take a simple concrete case, say x = 1. Then
\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}
What is the limit of this expression as n \rightarrow \infty?

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]

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?

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

 

[Felafel]

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

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]

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?

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]

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?

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]

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?

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]

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 :)

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.