Proving the Existence of Limits for Functions: A Study Guide

[╔(σ_σ)╝]

Homework Statement

Doing some studying for my midterm and came across these problems ...
a)
$f : D \rightarrow R$ with $a \leq f(x) \leq b$ for all c in D\{c}.

Show that if $lim_{x \rightarrow c} f(x)$ exist then $a \leq lim_{x \rightarrow c} f(x) \leq b$

b) Same thing except we have $g(x) \leq f(x) \leq h(x)$ and $lim_{x \rightarrow c} g(x) = lim_{x \rightarrow c} h(x) = L$

I need to show $lim_{x \rightarrow c} f(x)= L$.

The Attempt at a Solution

Is this as easy as I think or am I supposed to be more rigorous about the proof ?
a)
$$a - L \leq lim_{x \rightarrow c} f(x) -L \leq b - L$$

$$a - L \leq 0 \leq b - L$$

Thus,

$$a \leq L \leq b$$b) Same "proof" as in part 1.

:(

Will this suffice ?

 

[micromass]

To be honest, I don't really understand your proof.
How do you find $$a-L\leq \lim{f(x)}-L\leq b-L$$, isn't that what you need to show?

 

[╔(σ_σ)╝]

Well the question tells me ...
$$a \leq f(x) \leq b$$.

I subtracted L from all sides and took the limit of all sides. The question said that L exist.

 

[micromass]

Ah yes, but wasn't it easier to just take the limit of $$a\leq f(x)\leq b$$.

And you still need to show that if $$a\leq f(x)$$ then $$a\leq \lim{h(x)}$$, i.e. that limits preserve inequalities. I know this is trivial, but hey...

The same reasoning won't work for the second question, since you don't know there that $$\lim_{x\rightarrow c}{f(x)}$$ exists. If it existed, then it would equal L, but you don't know it exists yet... I fear an epsilon-delta is the only thing that can do the trick for question 2...

 

[╔(σ_σ)╝]

Ah yes, but wasn't it easier to just take the limit of $$a\leq f(x)\leq b$$.

And you still need to show that if $$a\leq f(x)$$ then $$a\leq \lim{h(x)}$$, i.e. that limits preserve inequalities. I know this is trivial, but hey...

The same reasoning won't work for the second question, since you don't know there that $$\lim_{x\rightarrow c}{f(x)}$$ exists. If it existed, then it would equal L, but you don't know it exists yet... I fear an epsilon-delta is the only thing that can do the trick for question 2...

XD it was late at night. You are right i should have just taken the limit of the inequality right away.

So proof 1 is okay, right ?

For proof two I took limit g(x) - L = limit of h(x) - L = 0

And i got $0 \leq lim f(x) - L \leq 0$

Thus,
lim f(x) = L.

 

[micromass]

But how do you know $$\lim{f(x)}-L$$ exists? You only know it is between h and g...

 

[╔(σ_σ)╝]

True.
I guess I will use my friends epsilon and delta.

I'll do that and get back to you l8tr.

:-)

 

[╔(σ_σ)╝]

I have a proof.

Let U be a neighbourhood of c and V be an epsilon neighbourhood of L.

Since when x is in U, g and h are in V and f is between g,h then f must be in V when x is in U.

Thus, lim f(x) = L.

I have a second proof btw.

How's this ?

 

[micromass]

That seems to be ok

 

[╔(σ_σ)╝]

Alright, thanks a bunch :-).