MHB What is the objective of proving the limit for max and min?

[Amad27]

Hello,

I am working towards an extremely difficult real analysis problem. The statement is as follows:

Prove that if $\lim_{x \to a} f(x) = l$ and $\lim_{x \to a} g(x) = m$ then $\lim_{x \to a} \max(f(x), g(x)) = \max(l, m)$ Some definitions:

$$\max(f, g)(x) = \frac{f + g + |g - f|}{2}$$
$$\max(l, m) = \frac{l + m + |m - l|}{2}$$

$$\lim_{{x}\to{a}} \max(f, g)(x) = \lim_{{x}\to{a}} \frac{f + g + |g - f|}{2}$$

$$= \frac{l + m}{2} + \lim_{{x}\to{a}} \frac{|g - f|}{2}$$

Somehow, the objective is to prove (using epsilon/delta) that

$$\lim_{x \to a} |g - f| = |m - l|$$

Let $H(x) = |g(x) - f(x)|$

$$\lim_{{x}\to{a}} H(x) = \lim_{{x}\to{a}} |g(x) - f(x)|$$

So prove that
$\displaystyle \left| |g(x) - f(x)| - Q \right| < \epsilon$ such that $|x - a| < \delta'$

And that $Q = |M - L|$

Can someone give me a hint, not the full solution?

 

[Opalg]

Hint: Use the reverse triangle inequality to show that $$\lim_{x\to a}(g(x) - f(x)) = m-l$$ implies $$\lim_{x\to a}|g(x) - f(x)| = |m-l|.$$

 

[Amad27]

Hint: Use the reverse triangle inequality to show that $$\lim_{x\to a}(g(x) - f(x)) = m-l$$ implies $$\lim_{x\to a}|g(x) - f(x)| = |m-l|.$$

Hello Opalg, thank you for the pleasant reply.

$$|g(x) - f(x) - (m-l)| < \epsilon$$

$$\left| |g(x) - f(x) - (m - l)| \right| < |g(x) - f(x) - (m - l)| < \epsilon$$

Wait... so what is the objective? Do we have to find a $\delta$ so that the statement is true?

Please answer this: In general, when asked for proofs like this, are we supposed to find a $\delta$ or is the presupposition that the limit already exists, so that it is already true that $|h(x) - Q| < \epsilon$ for $|x - a| < \delta'$??

Thanks!