Need help understanding limits, and their evaluation

1. Oct 8, 2012

gothloli

1. The problem statement, all variables and given/known data
I don't know how to prove that the lim $_{x-->a}$ f(x) =f(a)

2. Relevant equations
basically how do we know that the limit of the function near a point a, is f(a). My textbook (Calculus Michael Spivak, 4th ed.) says that we prove this using theorem 2, which is the properties of limits...

3. The attempt at a solution
I really don't understand, the textbook says we can prove this directly, but it doesn't show it

2. Oct 9, 2012

voko

You need to have a particular function, not generic f(x), or at least you need to know some properties of the function to prove anything about its limits.

3. Oct 9, 2012

HallsofIvy

Staff Emeritus
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 [itex]x\ne a[/tex] and f(a)= 2a" is continuous at x= a.

Last edited: Oct 9, 2012
4. Oct 9, 2012

gothloli

Thank you for the clear explanation