Here's what I'm trying to prove:
Let $f$ and $g$ be functions. Suppose that the following hold:
$f(x) \leq g(x)$
$\lim_{x \to x_0} f(x) = L$ and $\lim_{x \to x_0} g(x) = M$
Then, $L \leq M$
Proof Attempt:
By hypothesis, for any $\epsilon > 0$, there exist $\delta_1,\delta_2 > 0$ such that:
$0 < |x-x_0| < \delta_1 \implies |f(x) - L| < \epsilon$
$0 < |x-x_0| < \delta_2 \implies |g(x) - M| < \epsilon$
Let $\delta = \min\{\delta_1,\delta_2\}$. Then, if we have $0 < |x-x_0| < \delta$, we get:
$L-\epsilon < f(x) \leq g(x) < M + \epsilon$
$\implies 0 < (M-L) + 2\epsilon$
Suppose that $M-L < 0$. Let $\epsilon = \frac{L-M}{2}$. Then, this gives us:
$0 < (M-L) + (L-M) = 0$
That is an absurdity. Hence, $M-L \geq 0$ so $M \geq L$. That proves the desired result.
Does the proof above work? If it doesn't, why? How can I fix it?