Studying the behavior of f(x) = (-1)^x when x pass by number sequence

[bouhrassa]

Hello, I'm doing a calculus course from home and i'm struggling with a question. I'm french so I'll do my best to translate

a) Study the behavior of f(x) = (-1)^x when x pass by the number sequence : 1/3, 1/5, 1/7, 1/9, ...

b) 2/3, 2/5, 2/7, 2/9

c) What can you say about lim-->0 f(x) and f(0)

I'm just not sure what to talk about really. I was thinking maybe talk about the continuity of the function or something like that. If I could find the relation between a) and b), I'm confident that I could find the answer to c)

Thank you for any help.

 

[stapel]

a) Study the behavior of f(x) = (-1)^x when x pass by the number sequence : 1/3, 1/5, 1/7, 1/9, ...

When you say that x "passes by" the number sequence, do you perhaps mean that x takes on the values (one by one) of the numbers in the sequence? If so, then plug the given numbers in for the given variable in the given function. Note that fractional powers indicate radicals so, for instance, "1/2" indicates the square root and "1/3" indicates the cube root. What values are you getting for f(x)? What do you expect to see, as the sequence continues? (If not, please ignore what follows, and reply with clarification.)

b) 2/3, 2/5, 2/7, 2/9

Do the same as for part (a), but this time, you square. So, for instance, "2/3" means either the square of the cube root of -1, or else the cube root of the square of -1. What do you see? What do you expect to see, as the sequence continues?

c) What can you say about lim-->0 f(x) and f(0)

Use what you see, in parts (a) and (b), to say what you expect the limit value to be, or else state why there can't be a limit value. For the value of f(0), just plug in and evaluate. Compare the results of parts (a) and (b) to state something about continuity at x = 0 (remembering that "continuous" means "actually takes on the limit value".

If you get stuck, please reply showing your thoughts and efforts, based on the questions and instructions above. Thank you!

 

[bouhrassa]

I meant that x takes on the values of the numbers in the sequence as you said.

a) The way I see the first sequence is that it will always be equal to -1 since the cube root of -1 should always be equal to -1 if the root number is odd.

b) For the next sequence, if I square -1, I'll get +1. Then I'll apply the root as for the previous sequence but the answer for f(x) should alway be +1 for this sequence.

c) The limit value should be +1 I think since I can just state x=0 and f(0) = +1

I would say that the function is not continuous because it seems to jump from +1 to -1 but I'm really not sure about that.

Thanks for your help

P.S. sorry for the bad english.

 

[stapel]

a) The way I see the first sequence is that it will always be equal to -1...

c) The limit value should be +1 I think since I can just state x=0 and f(0) = +1

How can the limit value be +1 when you know that there exists a sequence of x-values which has f(x) always equal to -1? (Hint: Check the definition of "limit"!)

I would say that the function is not continuous because it seems to jump from +1 to -1 but I'm really not sure about that.

Again, check the definition of "continuous", especially as it relates to taking on the value of the limit.

 

[bouhrassa]

I'm not sure if I get this right. Sorry I don't have any teacher because I'm doing the course by myself.

The limit only exist if the value of lim-->0+ and 0- is the same.
The function is continuous if the limit exist.

I get 2 values for the same limit with 2 sequences.
If I can't get only one value for the limit, than it does not exist.
If I don't have a limit for x-->0, then the function isn't continuous at this point.

 

[stapel]

I'm not sure if I get this right. Sorry I don't have any teacher because I'm doing the course by myself.

The limit only exist if the value of lim-->0+ and 0- is the same.

Almost: The limit exists only if the value of the limit, by these or any other sequence of values approaching zero, is the same.

The function is continuous if the limit exist.

Almost: The function is continuous if the limit exists and if the function takes on that limit value at that x-value.

I get 2 values for the same limit with 2 sequences.

In particular, you get two different, conflicting, values for the two limit sequences.

If I can't get only one value for the limit, than it does not exist.

Correct.

If I don't have a limit for x-->0, then the function isn't continuous at this point.

Correct!

 

[bouhrassa]

A thousand thank you for your help !