Proving One-Sided Limits - Jack

[jackbauer]

Hi people,
could anyone tell me how to prove that the limit as f(x) approaches a from above equals the limit as f(x) approaches a from below? I can't see how to approach this proof, thx

Jack

 

[mathman]

The question as you stated it is too vague. In any case if f(x) is discontinuous, it just won't be true.

 

[Jameson]

Unless you have some sort of piecewise function, I see this as fairly straightfoward. If

(1) \lim_{x\rightarrow{a+}}f(x)=f(a)=\lim_{x\rightarrow{a-}}f(x),

then f(x) is continuous at x=a. What is your proof concerning? Continuity, delta-epsilon proofs?

 

[fourier jr]

i'm guessing it's the epsilon-delta stuff.

 

[Joffe]

Unless you have some sort of piecewise function, I see this as fairly straightfoward. If

(1) \lim_{x\rightarrow{a+}}f(x)=f(a)=\lim_{x\rightarrow{a-}}f(x),

then f(x) is continuous at x=a. What is your proof concerning? Continuity, delta-epsilon proofs?

That condition is not accurate, consider this function:

f(x) = \frac{x^2-1}{x-1}

The limit above and below f(1) is equal to 2 though it is undefined at that point.

 

[Muzza]

It's obvious that Jameson meant for f(a) (a = 1 in this case) to exist, seeing as he mentioned that something should be equal to it, and in that case, the condition is accurate.

 

[Robokapp]

I know what he means. He is talking about a function between points (a,f(a)) and (b,f(b)) and he wants to know how to prove the limit at x=a or x=b

From what I remember in Chapter 2, Calculus AB all you need to do is see limit as x->a or b->b from the existent side, and then plug in the value into the function. If it's the same, it's continuous. If it's not, no continuity.

It's like the following:

limit as x->a of f(x)=b
and f(a)=b

makes a function continuous at point (a,f(a))

right? It's been a good few months.

 

[Jameson]

Yes, I was stating the conditions for a function being continuous at the point a. However, I was not saying continuity is necessary for a limit to exist. Sorry if I was unclear.