# Help with a continuous function lemma

CuriousBanker
I am on page 97 of Spivak calculus and having trouble proving Theorem 3 of chapter 6 (which he says is a lemma for the next chapter). I don't know how to type the symbol delta so I am replacing delta with @, and replacing epsilon with &

Theorem: Suppose f is continuous at a, and f(a)>0. Then there is a number @>0 such that f(x)>0 for all x satisfying |x-a|<@

Proof: Consider the case f(a)>0 Since f is continuous at a, if &>0 there is a @>0 such that, for all x, if |x-a|<@, then |f(x)-f(a)<&

Since f(a)>0, we can take f(a) as & Thus there is @>0 so that for all x, if |x-a|<@ then |f(x)-f(a)|<f(a)

So far, I understand all of this. The part I don't understand is the very last line of the proof, which says:

and this last inequality implies f(x)>0

I don't get how if |x-a|<@ then |f(x)-f(a)|<f(a) implies that f(x)>0.

INTUITIVELY I get it...because if |f(x)-f(a)|<f(a) that means the distance between f(x)-f(a) is smaller than the number f(a), and f(a)=f(a)-0, so if f(x) were negative, the distance between f(x) and f(a) would be greater than the distance between f(a) and 0. I get it...but I jus't cant prove it with rigor.

Thank you in advance for your help

Somebody posted this answer on yahoo answers:

If |f(x) - f(a)| < f(a), then -f(a) < f(x) - f(a) < f(a), so in particular -f(a) < f(x) - f(a). Adding f(a) to both sides of this inequality gives 0 < f(x).

But I dont see why the absolute value signs got dropped

## Answers and Replies

Homework Helper
If you want, you can go back to the definition of the absolute value, which is
$$|z| = \begin{cases} -z & \text{ if } z < 0; \\ 0 & \text{ if } z = 0; \\ z & \text{ if } z > 0. \end{cases}$$
(actually you can cover the = 0 case in one of the inequalities, but let's be explicit).

So let's work through them, setting z = f(x) - f(a):
(1) The trivial case is f(x) - f(a) = 0, so |f(x) - f(a)| < f(a) immediately gives you 0 < f(a).
(2) if f(x) - f(a) < 0 then you get |f(x) - f(a) = -(f(x) - f(a)) < f(a). [...]
(3) if f(x) - f(a) > 0 then you get |f(x) - f(a)| = f(x) - f(a) < f(a). [...]
Can you fill the dots by yourself now?

CuriousBanker
Thanks for the reply, I have been busy all week so I just noticed it.

Case 1 I see how you got 0 < f(a), but I don't see how that implies f(x)>0. Again I understand it intuitively but I still don't see how it is proven by this case
Case 2 I filled in the dots, and it makes complete sense to me
Case 3 I filled in the dots, and it makes complete sense to me

R136a1
Case 1 I see how you got 0 < f(a), but I don't see how that implies f(x)>0. Again I understand it intuitively but I still don't see how it is proven by this case

Because by assumption, ##f(x)-f(a)=0## and thus ##f(x)=f(a)##.

CuriousBanker
Wow I am slow :)

Thanks for pointing that out

CuriousBanker
One more question....so the epsilon-delta limit definition states that for ALL epsilon>0, there is some delta>0 such that for all x if 0<|x-a|<delta, then |f(x)-L|<epsilon. The key word being ALL.

In the example spivak game he assumed Epsilon=f(a). What about for all other epsilon? How does proving that it works when epsilon=f(a) prove that it works for all epsilon?

Staff Emeritus
Gold Member
2021 Award
You're told that the function is continuous, you never have to prove that any limits exist at any point in time.

CuriousBanker
You're told that the function is continuous, you never have to prove that any limits exist at any point in time.

I'm not sure I understand. if we don't have to prove any limits exist at any point in time why is the whole proof involving proving a limit?

What's the point of proving it works only when Epsilon = f(a) if we cannot prove it works given any epsilon? What if epsilon does not equal f(a)?

Staff Emeritus
$$\lim_{x\to a} f(x) = a$$.