Proving that $L \le M$ as limits of $f$ and $g$ when $f(x) \le g(x)$ While doing some tasks for my next calculus course, I ran across this task:
"Let $a < b <c$, and assume that $f(x) \le g(x)$ for all $x \in [a, c]$. If $\lim_{x \to b}f(x) = L$ and $\lim_{x \to b}g(x) = M$, prove that $L \le M$."
I have done some thinking about this, but I have to admit that I'm hopelessly stuck. A proof by contradiction seems natural here, so I'm assuming, for the sake of contradiciton, that $L > M$. From the limits we know, from the definition, that $|x - b| < \delta(\epsilon)$ and that $|f(x) - L| < \epsilon$ and $|g(x) - M| < \epsilon$. Unconventionally, I tried setting $\epsilon < 1$ (yes, $\epsilon$, not $\delta$), getting $|f(x) - L| < 1$ and $|g(x) - M| < 1$. I'm having a hard time forcing a contradiction from this. We are left with the facts that $f(x) < 1 + L$ and $g(x) < 1 + M$. 
While writing this, I thought that if I changed $\epsilon < 1$ to "very small $\epsilon$", we could force a contradiction from $f(x) < \epsilon + L$ and $g(x) < \epsilon + M$. For sufficiently small $\epsilon$ and $\delta(\epsilon)$, this would imply (I do not have any confidence in this statement whatsoever) $f(x) > g(x)$ for $L > M$, which is the desired contradiction, and we are done.
I'm new to these kind of proofs, and I'd highly appreciate to be spoonfed about my mistakes here.
 A: You started very well. 
To finish the proof, instead of setting $\epsilon=1$, try
$$\epsilon=\frac{L-M}{3}$$
Also, note that
$$|f(x) - L| < \frac{L-M}{3} \Leftrightarrow -\frac{L-M}{3} < f(x) - L < \frac{L-M}{3}$$
$$|g(x) - M| < \frac{L-M}{3} \Leftrightarrow -\frac{L-M}{3} < g(x) - M < \frac{L-M}{3}$$
Now combine these two inequalities with
$$g(x) -f(x) \geq 0$$
Note Intuitively, the proof reduces to the following: If $L>M$, as $f$ gets very close to $L$ and $g$ gets very close to $M$, we cannot have $g \geq f$. What we did, we split the interval $[M,L]$ in three thirds, and made sure that $f$ is in the top third, while $g$ is in the bottom third.
A: It suffices to prove that $\lim_{x\to b}f(x)=\ell\ge0$ if $f(x)\ge 0$.
Hint By definition we have
$$\ell-\epsilon<f(x)<\ell+\epsilon\quad \text{if}\; |x-b|<\delta$$
so by contradiction take $\epsilon=-\frac\ell 2$.
A: You need to go "arbitrary small", otherwise you won't get the statement. 
Suppose $L > M$, and call $\varepsilon = L-M > 0$. Then there exists $\delta > 0$ such that $0 < |x - b| < \delta$ implies $|f(x) - L| < L-M$. This means in particular that $L - f(x) < L - M$, i.e. $f(x) > M$. Also, there exists $\delta' > 0$ such that $0 < |x-b| < \delta'$ implies $|g(x) - M| < L-M$, which implies $g(x) < L$. Combining this together gives
$$
M < f(x) \le g(x) < L < M
$$
for $|x - b| < \min \{ \delta, \delta' \}$, a contradiction.
Hope that helps,
