Study the continuity of this function

In summary: You can't say that the function is not continuos at x<0 because that would mean that it doesn't exist.
  • #1
Felafel
171
0

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
 
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  • #2
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]?
 
  • #3
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
 
  • #4
Felafel said:
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.
 
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  • #5
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..
 
  • #6
Felafel said:
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
[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]?
 
  • #7
jbunniii said:
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]?

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?
 
  • #8
Felafel said:
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)?
 
  • #9
Dick said:
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?
 
  • #10
Felafel said:
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.
 
  • #11
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?
 
  • #12
Felafel said:
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.
 
  • #13
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?
 
  • #14
Felafel said:
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?
 
  • #15
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 :)
 
  • #16
Felafel said:
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.
 

What is the definition of continuity for a function?

The definition of continuity for a function is the ability for a function to have a smooth and uninterrupted graph without any holes, breaks, or jumps. This means that the function can be drawn without lifting the pen from the paper.

How can I determine if a function is continuous?

To determine if a function is continuous, you can check if it satisfies the three criteria for continuity: 1) the function is defined at the point, 2) the limit of the function at the point exists, and 3) the limit and the function value at the point are equal.

What is the importance of studying continuity of a function?

Studying continuity of a function is important because it helps us understand the behavior of a function and predict its values. It also allows us to identify and analyze any points of discontinuity, which may have significant implications in practical applications.

What are the different types of discontinuities in a function?

There are three types of discontinuities in a function: removable, jump, and infinite. A removable discontinuity occurs when there is a hole in the graph of the function, a jump discontinuity occurs when there is a jump or gap in the graph, and an infinite discontinuity occurs when the function approaches infinity at a certain point.

How can I improve my understanding of continuity in functions?

To improve your understanding of continuity in functions, you can practice solving problems and graphing different types of functions. It may also be helpful to review the properties and theorems related to continuity, such as the Intermediate Value Theorem and the Squeeze Theorem.

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