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

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 by a moderator: Oct 9, 2012
4. Oct 9, 2012

gothloli

Thank you for the clear explanation

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