I know I've posted quite a few questions about epsilon-delta proofs in the past few days, but I really do appreciate all the help Math Help Forum has given me so far, I think I'm getting better at it. Now, the following problem, I beleive I've come close to solving. But, as always, I really need math forum to help verify the valididty of the proof. Heres the problem exactly:

Prove that if:

$\displaystyle \lim_{x \to a} f(x) = L$

and

$\displaystyle \lim_{x \to a} g(x) = M$

then

$\displaystyle \lim_{x \to a} max( \; f(x), \; g(x) \ = max(\; L, \; M\$

Now, I will post the work I've come up with so far, and hopefully any mistakes or major errors will be found by the every diligent members of Math Help Forum. But, my supposed proof uses a property of limits I've proved earlier in this textbook, so I'll place it immediately below and call it "property [A]":

Now, back to the main problem, heres the work I've come up with so far. Lets look at seperate cases that need to be adressed:PROPERTY [A]

If, for all $\displaystyle x$, the following holds:

$\displaystyle f(x) \leq g(x)$

then it follows that:

$\displaystyle \lim_{x \to a} f(x) \leq \lim_{x \to a} g(x)$

given that these limits exist.

CASE 1:

Suppose that:

$\displaystyle f(x) < g(x)$

for all $\displaystyle x$. Then, it follows from [A] that:

$\displaystyle L < M$

and from the above we get:

$\displaystyle \lim_{x \to a} max( \; f(x), \; g(x) \; ) = \lim_{x \to a} g(x) = M = max(\; L, \; M \$

Now, I understand that there is a difference between $\displaystyle a < b$ and $\displaystyle a \leq b$, so I am not sure if the little difference I made invalidates the use of property [A]. I figured it only did not use it in full strength, since the way I stated it did not allow for the possiblity $\displaystyle f(x) = g(x)$, so that must be a thrid Case, correct? I'm not sure how to prove Case 3??? This is where I need some help. Its mostly do to the fact that I am confused as to what the following expression would mean, intuitvely and in terms of the proof:

CASE 2:

Suppose that:

$\displaystyle g(x) < f(x)$

then, by [A] it follows simmilarly (like Case 1) that:

$\displaystyle M < L$

and from the above we get:

$\displaystyle \lim_{x \to a} max( \; f(x), \; g(x) \; ) = \lim_{x \to a} f(x) = L = max(\; L, \; M \$

$\displaystyle max(\; a, \; b \$

if

$\displaystyle a=b$

Also, I don't know if the above methodology towards a proof is valid either. To be completely rigourous, is it neccesarry to deconstuct this problem into the language of the definition of a limit? (i.e. epsilon and delta inequalities?)

Thanks in advance for any help.