Proving limit properties with epsilon-delta definitions

[gimpy]

Hi, I am 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 I am 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.

 

[Hurkyl]

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!

 

[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...

 

[Hurkyl]

$$\lim_{x\rightarrow a} h(x) = \lim_{x\rightarrow a} g(x) - \lim_{x\rightarrow a} f(x) \geq 0$$

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.

 

[gimpy]

Originally posted by Hurkyl
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.

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 ;)

 

[Hurkyl]

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.