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: 
The hypothesis is that $f(x)\le g(x)$ for all $x$. We want to show under this hypothesis that

$$\lim\_{x\to a}f(x)\le\lim\_{x\to a}g(x)\;.\tag{1}$$

The idea of the proof is to show that if $(1)$ is false, then so is the hypothesis: we’ll assume that $(1)$ is false and use that assumption to find an $x$ such that $f(x)\not\le g(x)$, i.e., such that $f(x)>g(x)$.

The first step is just for convenience, so that we don’t have to write so much: we let $\ell=\lim\_{x\to a}f(x)$ and $m=\lim\_{x\to a}g(x)$. Now we assume that $(1)$ is false, i.e., that $\ell>m$. (You have $\ell\ge m$ in your question, but that can’t be right.) Then we let $\epsilon=\ell-m$; since we’re assuming that $\ell>m$, we know that $\epsilon>0$.

Now we apply the definition of limit twice. Since $\ell=\lim\_{x\to a}f(x)$, there is a $\delta\_f>0$ such that $|\ell-f(x)|<\frac{\epsilon}2$ whenever $0<|x-a|<\delta\_f$. And since $m=\lim\_{x\to a}g(x)$, there is a $\delta\_g>0$ such that $|m-g(x)|<\frac{\epsilon}2$ whenever $0<|x-a|<\delta\_g$. Now let $\delta=\min\{\delta\_f,\delta\_g\}$; clearly

$$|\ell-f(x)|<\frac{\epsilon}2\quad\textbf{and}\quad|m-g(x)|<\frac{\epsilon}2$$

whenever $0<|x-a|<\delta$. The first of these implies that

$$f(x)>\ell-\frac{\epsilon}2$$

whenever $0<|x-a|<\delta$, and the second implies that

$$g(x)<m+\frac{\epsilon}2$$

whenever $0<|x-a|<\delta$. Put the two together: if $0<|x-a|<\delta$, then

$$g(x)<m+\frac{\epsilon}2=\ell-\frac{\epsilon}2<f(x)\;.$$

We’ve found not just *one* $x$ such that $g(x)>g(x)$: we’ve found a whole interval of them. This certainly contradicts the hyposthesis that $f(x)\le g(x)$ for all $x$, and that contradiction shows that our assumption that $\ell>m$ is impossible. Thus, it must be true that $\ell\le m$, which is what we wanted to prove.
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.