Saying that "\lim_{x\to a} f(x)= f(a)" is the definition of "continuous at x= a" and the great majority of functions are NOT continuous so you certainly cannot prove that for all f.
How you would prove it for a specific function depends on the function, as well as what limit theorems you have learned.
For example, to prove that f(x)= 3x- 4 is continuous for all x, that is, to prove that \lim_{x\to a} 3x- 4= 3a- 4, directly from the definition, I would look at the definition of "\lim_{x\to a} f(x)" and see that it involves "|f(x)- L|< \epsilon". Here, f(x)= 3x- 4 and L, the limit I want to prove, is f(a)= 3a- 4. So I know that I need to look at |(3x-4)- (3a- 4)|= |3x- 3a|= 3|x- a|. And I need to prove that 3|x- a|< \epsilon as long as |x- a|< \delta where I need to be able to choose a suitable \delta for any given \epsilon. Comparing "3|x-a|< \epsilon" and "|x- a|< \delta" it immediately occurs to me that I need to choose \delta= \epsilon/3 (or, strictly speaking any smaller number).
But if I have the "limit theorems" that
1) \lim_{x\to a} C= C for any constant C
2) \lim_{x\to a} x= a
(those are often called the "trivial limits")
3) if \lim_{x\to a} f(x)= F and ]\lim_{x\to a} g(x)= G, then \lim_{x\to a} (f+ g)(x)= F+ G
4) if \lim_{x\to a} f(x)= F and ]\lim_{x\to a} g(x)= G, then \lim_{x\to a} (fg)(x)= FG
5) if \lim_{x\to a} f(x)= F and ]\lim_{x\to a} g(x)= G, AND G\ne 0, then \lim_{x\to a} (f/g)(x)= F/G
I can immediately say that \lim_{x\to a} 3x- 4= (lim_{x\to a}3)(\lim_{x\to a}x)+ (\lim_{x\to a} -4)= 3a- 4
If I also know the very important but often overlooked limit theorem
6) If f(x)= g(x) for all x close to but NOT equal to a, then \lim_{x\to a}f(x)= \lim_{x\to a} g(x)
I can argue that, for x not equal to a, (x^2- a^2)(x- a)= x+ a so that \lim_{x\to a}(x^2- a^2)/(x- a)= \lim_{x\to a} x+ a= 2a. (Which I could NOT do using (5) because the denominator goes to 0.)
So that I have proved that the function "f(x)= (x^2- a^2)/(x- a) as long as x\ne a[/tex] and f(a)= 2a" is continuous at x= a.
