# Homework Help: Does this limit exist?

1. Jun 25, 2013

### Andrax

This has confused me a bit
I'm typing on my phone so I can't use math symbols sorry
F(x) = 1 ; x=5
0 otherwise (x=/5)
Does lim X approaches 5 f(x) exists and = 1(f continuous at 5) ? Or does it not(because when we approach a were also in the x=/a space so there is also 1)
can someone be really kind and Prove it using epsilon?
I know you can just put alpha = any number you want since 0<epsilon always
I think I'm confusing these functions with the ones like sinx..

Last edited: Jun 25, 2013
2. Jun 25, 2013

### haruspex

Not at all sure what you mean by that last statement. When x ≠ a, F(x) = 0, yes? What is F(1/n), n = 1, 2, ...?

3. Jun 25, 2013

### Andrax

I'm sorry forgot to set the integer fixed

4. Jun 25, 2013

### Andrax

F(x) : R--) R
0 x=5
1 x=/5
Is f continuous at 0?

5. Jun 25, 2013

### Ray Vickson

What is the *definition* for f(x) to be continuous at x = 0? Does your f satisfy that definition? Show your work!

6. Jun 25, 2013

### lurflurf

The point of this exercise is to make the point that the value of a function at the value of the limit does not matter.
That is

$$\lim _{x \rightarrow a} \mathop{f}(x)$$
Does not depend on the value of f(a)

Perhaps there is a function G that you know of such that
G(x)=A if x=5
G(x)=F(x) otherwise (x=/5)
Choose G(5) whatever you like. So that you know the limit.

7. Jun 26, 2013

### Andrax

I'm sorry but I had a huge confusion with this, I'm not sure if when it approaches 5 is it f(5) or not..
I used epsilon delta but. I can't decide what f(x) should I be using
lx-5l<alpha. lf(x)-f(5)l< epsilon..
What f(x) should I use
Just to mention this is not my first time with limits but out of nowhere I found myself unable to do this..

8. Jun 26, 2013

### Andrax

What are you saying is that the limit of my function x approaches 5 is. 1 and not 0?
We can choose g(5)=100 that doesn't mean limit x approaches 5 is 100?

9. Jun 26, 2013

### Andrax

thanks everyone i've got it the limit = 1 => f is not continious at 5
this whole problem was caused by an exercise @ spivak's calculus it was f is discon tinious at l but i thought it is continious .