Thread: Determine where f has a limit

Define f:R--->R as follows:
f(x)=x-[x] if [x] is even
f(x)=x-[x+1] if [x] is odd

Determine the points where f has a limit and justify

Not quite sure how to get started...

Originally Posted by kathrynmath

Determine the points where f has a limit and justify. Not quite sure how to get started...

Just draw a graph.

So, I just use the graph and see where ther is a limit? Then I can use a formula to justify that these are correct limits?

Originally Posted by kathrynmath

So, I just use the graph and see where ther is a limit? Then I can use a formula to justify that these are correct limits?

YES And here is a graph.

Ok, so there is a limit everywhere except for the integers. how would I justify using a theorem. Lim(F(x)+G(x))=LimF(x)+LimG(x)?

Originally Posted by kathrynmath

Ok, so there is a limit everywhere except for the integers. how would I justify using a theorem. Lim(F(x)+G(x))=LimF(x)+LimG(x)?

Never mind I got the first part justified. then for the 2nd part, i think I find 2 sequences that converge to the same limit, but f(x_n) doesn't converge to the same limit. hence limit DNE.

Originally Posted by kathrynmath

Ok, so there is a limit everywhere except for the integers. how would I justify using a theorem. Lim(F(x)+G(x))=LimF(x)+LimG(x)?

Is there a limit everywhere except at the integers. Do I consider each case seperately where [x} is even and the other case?

Limit only at the even numbers?
if 1<x<2, [x]=1 Then f(x)=x-2. The limit=0
If 2<x<3 [x]=2 Then f(x)=x-2 The limit =0

a-1<x<a where a is an odd number. [x]=a-1. Then f(x)=x-(a-1). Limit as x goes to a=1
a<x<a+1 where a is an odd number. [x]=a. Then f(x)=x-(a-1+1). Limit as x goes to a= 0
a-1<x<a where a is an even number. [x]=a-1. Then f(x)=x-(a-1+1). Limit=0
a<x<a+1 where a is an even number. [x]=a. Then f(x)=x-a. Limit = 0

So the limit exists for even numbers, it seems.

Originally Posted by kathrynmath

Limit only at the even numbers?
if 1<x<2, [x]=1 Then f(x)=x-2. The limit=0
If 2<x<3 [x]=2 Then f(x)=x-2 The limit =0

a-1<x<a where a is an odd number. [x]=a-1. Then f(x)=x-(a-1). Limit as x goes to a=1
a<x<a+1 where a is an odd number. [x]=a. Then f(x)=x-(a-1+1). Limit as x goes to a= 0
a-1<x<a where a is an even number. [x]=a-1. Then f(x)=x-(a-1+1). Limit=0
a<x<a+1 where a is an even number. [x]=a. Then f(x)=x-a. Limit = 0

So the limit exists for even numbers, it seems.

Is this the right idea?