# Analysis Proof Help :S

1. Jan 25, 2004

### gimpy

Hi, im taking my first analysis course and we are studying Limits right now. My prof said they are the most important thing to remember out of this whole semester. Anyways i have two problems im trying to solve that i could do with some help.

1) Show that if $$f(x) \leq 0$$ and $$\lim_{x->a} f(x) = l$$, then $$l \leq 0$$.

2) If $$f(x) \leq g(x)$$ for all x, then $$\lim_{x->a} f(x) \leq \lim_{x->a} g(x)$$.
If those limits exist.

For number one i can see this is obvious but i don't know where to start to try and prove it.

I know the definitions for limits, do i use them somehow?

For number 1) i think that i can use a proof by contradiction somehow.

2. Jan 25, 2004

### Hurkyl

Staff Emeritus
Yah, using the definition of limit will probably be helpful. Proof by contradiction sounds like a good plan, let us know how far you get!

3. Jan 25, 2004

### gimpy

I have been working ont he second question:

2) If $$f(x) \leq g(x)$$ for all x, then $$\lim_{x->a} f(x) \leq \lim_{x->a} g(x)$$.
If those limits exist.

Suppose that $$f(x) \leq g(x)$$. Let $$h(x) = g(x)-f(x)$$. So $$h(x) \geq 0$$.
$$\lim_{x->a} h(x) = \lim_{x->a} g(x) - \lim_{x->a} f(x) \geq 0$$. Therefore $$\lim_{x->a} f(x) \leq \lim_{x->a} g(x)$$

Is this right?

Can someone give me a hint for question 1?
Start me off or something......

Last edited: Jan 25, 2004
4. Jan 25, 2004

### Hurkyl

Staff Emeritus
What is your justification for this?

Your hint for problem (1) is to assume $l > 0$ and rewrite the entire problem in terms of epsilons and deltas; exactly what you said you thought you should do.

5. Jan 26, 2004

### gimpy

Ok i had a good nights sleep went to school and came back tonight with a fresh mind to give it another shot. And i believe it worked :D

1) Show that if $$f(x) \leq 0$$ and $$\lim_{x->a} f(x) = l$$, then $$l \leq 0$$.

$$\forall\epsilon > 0, \exists\delta > 0 \ni$$ for all $$x$$ , $$0< |x - a| < \delta$$, then $$|f(x) - l| < \epsilon$$.

Assume that $$l > 0$$

$$\forall\epsilon > 0, \exists\delta > 0 \ni$$ for all $$x$$ , $$0< |x - a| < \delta$$, then $$l - f(x) < \epsilon$$.

since $$l - f(x) > 0 \implies 0 < l - f(x) < \epsilon$$

add $$f(x)$$ to each side to get $$f(x) < l < \epsilon + f(x)$$.

Choose $$\alpha$$ to be the distance from $$f(x)$$ to $$0$$. Then we choose $$\delta \ni \epsilon = \alpha$$, then $$\epsilon + f(x) = 0$$
therefore, $$f(x) < l < 0$$ which is a contradiction to $$l > 0$$.

Im sure this must be correct.

I will try number 2 now. Thanks for your hint ;)

6. Jan 26, 2004

### Hurkyl

Staff Emeritus
You're almost there: given the assumption $l > 0$, you've correctly proven

$$\forall \epsilon > 0 \exists \delta > 0 \forall x: 0 < |x - a| < \delta \implies f(x) < l < f(x) + \epsilon$$

however, the next steps are wrong. For instance, the distance from $f(x)$ to zero could, in fact, be zero. (e.g. take $f(x) = 0$, or $f(x) = -|x|$)

I'm also not sure what you're trying to do when you say "choose $\delta$ such that $\epsilon = \alpha$"... when deriving the contradiction you do get to choose $\epsilon$, but the contradiction must hold for all $\delta$.

You need another hint on this one?

As for number two, the proof you gave almost works, now that you've proven #1... see if you can rewrite it to take advantage of knowing #1.