Question: Suppose that $f(x) \leq g(x) \text{ for all }x$. Prove that $\lim\_{x\to a} f(x) \leq \lim\_{x \to a} g(x)$ provided these limits exist.

Attempt: I am not even sure how to start this question. I had made an attempt but it was getting nowhere when I did it. I found a solution to the question, but I still do not understand what is going on:

Suppose $$ l = \lim\_{x\to a}f(x) \geq \lim\_{x\to a}g(x) = m$$. Let $$\epsilon = l - m > 0$$ 

Then there is $$\delta > 0 \text{ such that if } 0 <|x-a|<\delta \Rightarrow |l-f(x)|< \frac{\epsilon}{2} \text{ and } |m-g(x)| < \frac{\epsilon}{2}$$ 

Thus for $$0 <|x-a|<\delta \text{ we have } g(x) < m + \frac{\epsilon}{2} = l - \frac{\epsilon}{2} < f(x)$$ 

Contradicting the hypothesis.

My problem is I do not see clearly what the hypothesis is and of equal importance I don't see how this proof accomplishes the objectives of the question.

Answer: 
If you understand the meaning of limit clearly you will observe that the result in question is too obvious. If $f(x) \to l$ as $x \to a$ and $g(x) \to m$ as $x \to a$ then we can get the values of $f(x)$ as close to $l$ as we want by taking values of $x$ sufficiently close to $a$ and we can get values of $g(x)$ as close to $m$ as we want by taking $x$ sufficiently close to $a$.

Now let $l > m$. Since it is possible to make $f(x)$ close to $l$ and $g(x)$ close $m$ and $l > m$ then you will get values of $f(x)$ which are greater than $g(x)$ and hence $f(x) > g(x)$ which contradicts our hypotheses. Thus $l \leq m$.

The reasoning in the last paragraph depends on the following very very obvious fact: If $l > m$ then there is a neighborhood $A$ of $l$ and a neighborhood $B$ of $m$ such that all points of $A$ are greater than all points of $B$. Informally if $l > m$ then points near $l$ are greater than point nears $m$.

The use of $\epsilon - \delta$ type argument does not add any extra rigor to whatever I wrote above, it only adds formalism and makes the argument look high brow and perhaps is a good way to intimidate beginners. In fact this is precisely the reason *you did not understand what is going on* in the solution you found.
Your proof by contradiction should begin with a strict inequality: "Suppose $l = \lim\_{x\to a}f(x) > \lim\_{x\to a}g(x) = m$." Continuing:

Pick an $\epsilon > 0$ so small that the neighborhoods $Y\_{l} = ]l - \epsilon, l+\epsilon[, Y\_{m} = ]m - \epsilon, m+\epsilon[$ have no point in common. (**Note**: $l - \epsilon > m + \epsilon$.) There exist: (1) a neighborhood $X\_{f}$ of $a$ that gets mapped by $f$ into $Y\_{l}$, and (2) a neighborhood $X\_{g}$ of $a$ that gets mapped by $g$ into $Y\_{m}$. Therefore, every point $x$ in $X\_{f} \cap X\_{g}$ gets mapped into $Y\_{l}$ by $f$ and into $Y\_{m}$ by $g$. But then, by the **Note** above, $f(x) > g(x)$. This is a contradiction.