# Limit of f(x) as it approaches from the positive side

1. Dec 10, 2008

### tomboi03

i. Write out the proper definition for
lim f(x)
x$$\rightarrow$$a+

ii. Let g(x)=x-1cos(x-1) Show that for every N and any b>0 there is a point t E (0,6) so that g(t)>N

iii. Prove that for g as in (ii) we don not have
lim g(x)
x$$\rightarrow$$0+

I'm not sure what how to do any of these... like.. for the first one, i know that the definition of a limit is for every $$\epsilon$$>0 there is some $$\delta$$ >0 such that, for all x if 0<|x-a|< $$\delta$$, then |f(x)-L| < $$\epsilon$$

2. Dec 10, 2008

### Dick

Continue with the first one. All you have to do is drop the absolute value in the delta part. Do you see why that works?

3. Dec 10, 2008

### tomboi03

I can see why you drop the absolute value sign, it then becomes x>a meaning the x's approaching from the positive side.

so what about the others?

Thanks

4. Dec 10, 2008

### Dick

It's time for you to help. If t is in (0,6) then as t->0, 1/t becomes very large, right? What does cos(1/t) do?

5. Dec 10, 2008

### tomboi03

wouldn't it be either 1 or 0 or -1?

6. Dec 10, 2008

### Dick

It oscillates between 1 and -1, yes. So as 1/t becomes very large...?

7. Dec 10, 2008

### tomboi03

does it go to zero?

8. Dec 10, 2008

### Dick

No, (1/t) becoming very big times something oscillating between -1 and 1, will, I think oscillate between -infintity and +inifinity. You don't agree, right?

9. Dec 10, 2008

### tomboi03

i'm not following you. please elaborate.

10. Dec 14, 2008

### tomboi03

can someone please help me out with this problem.. the person that replied to this thread never fully answered my question

11. Dec 15, 2008

### Staff: Mentor

As Dick said earlier, it's time for you to do some work. He was trying to explain what the function did without spoonfeeding you the answer. I think he lost patience.

Go back and look at what he said. If you have specific questions, ask them and someone will help you out.

12. Dec 15, 2008

### Dick

I didn't lose patience (yet). But sketch a graph of the function. To help answer the given question think about the values of the function g(x) at x=1/(k*pi) where k is an integer.

13. Dec 15, 2008

### tomboi03

o i see, as it approaches zero, the graph starts to oscillate from negative infinity to positive infinity. But i'm not sure how to answer part ii?

14. Dec 15, 2008

### Dick

Try and answer the question I asked, what is g(1/(k*pi))?

15. Dec 15, 2008

### tomboi03

well, as k increases for g(1/k*pi)....
it is negative infinity for k=1
it is positive infinity for k=2
and etc...

16. Dec 15, 2008

### Dick

Why would that be? If x=1/(k*pi) what is cos(1/x)? k is an integer. It's not some kind of infinity.

17. Dec 15, 2008

### tomboi03

well, i was referring to the actual function because the actual function is x^(-1)cos(1/x) soooo, as k increases... g(1/k*pi) is negative infinity for k=1 and positive infinity for k=2... no?

18. Dec 15, 2008

### Dick

Negative infinity? If k=1? How does that follow? I get pi*cos(pi). That's not infinity. Express the answer in terms of k.

19. Dec 15, 2008

### tomboi03

o i see, when k=1, pi*cos(pi)

i'm sorry, i haven't gotten sleep for the passed few days...

soooo, in terms of k... it would be....
kpi*cos(kpi) right?

20. Dec 15, 2008

### Dick

Yes, cos(pi)=(-1), cos(2pi)=(+1) etc. You can also write that as kpi*(-1)^k. Does that help with question ii)? If that seems hard you are probably better off getting some sleep than trying to push on.