Proving that if $f(x) \le g(x)$ then $f(x_0) \le g(x_0)$ Suppose $f: D \to \mathbb R$ and $g:D \to \mathbb R$, $x_0$ is an accumulation point of $D$, and $f$ and $g$ have limits at $x_0$. If $f(x) \le g(x)$ for all $x \in D$, then 
$\lim \limits_{x \to x_0}f(x) \le \lim \limits_{x \to x_0}g(x)$
Assume that $\forall x \in D$,   $f(x)$ and $g(x)$ both have limits at some point $x_0$ and $f(x) \le g(x)$  
We know that $\lim \limits_{x \to x_0}f(x) = f(x_0)$ and $\lim \limits_{x \to x_0}g(x)=g(x_0)$
$\therefore f(x_0) \le g(x_0)$ because $f(x) \le g(x) $  $\forall x \in D$
 A: The two answers already posted do provide enough detail as to how a correct proof would go. I am puzzled why a bounty is put on this question, asking for more detail. I will try to provide more comments related to proposed proof by OP. 
First off, the question does not ask about $f(x_0)$ and $g(x_0)$, it is not assumed that $f(x_0)$ and $g(x_0)$ are defined. In other words, it is not assumed that $x_0\in D$. If $x_0$ were a member of $D$, then there is nothing to prove since it is given that $f(x) \le g(x)$ for all $x\in D$, hence in particular for $x_0$. But as we are not given that $x_0$ is in $D$ we should be prepared for the worse, and should not assume that $x_0\in D$. 
Therefore any reference to $f(x_0)$ and $g(x_0)$ is meaningless. 
Example to illustrate the above comments, as well as the statement of the problem. Let $p(x)=\dfrac{x^2+x-2}{x-1}$ and $q(x)=\dfrac{x^2+2x-3}{x-1}$. 
The domain of $p$ and $q$ is $(-\infty,1)\cup(1,\infty)$. We have $p(x)<q(x)$ for all $x$ in $(-\infty,1)\cup(1,\infty)$ since $q(x)-p(x)=\dfrac{x^2+2x-3-(x^2+x-2)}{x-1}=\dfrac{x-1}{x-1}=1>0$. Here $x_0=1$ and $p(x_0)$ and $q(x_0)$, that is, $p(1)$ and $q(1)$ are undefined. But we could  find the limits, $\lim\limits_{x\to1}p(x)=\lim\limits_{x\to1}\dfrac{x^2+x-2}{x-1}=\lim\limits_{x\to1}\dfrac{(x+2)(x-1)}{x-1}=\lim\limits_{x\to1}(x+2)=1+2=3$, and
$\lim\limits_{x\to1}q(x)=\lim\limits_{x\to1}\dfrac{x^2+2x-3}{x-1}=\lim\limits_{x\to1}\dfrac{(x+3)(x-1)}{x-1}=\lim\limits_{x\to1}(x+3)=1+3=4$.
So indeed $\lim\limits_{x\to1}p(x)\le\lim\limits_{x\to1}q(x)$, as $3\le4$. 
Concerning the statement by OP that: "We know $\lim \limits_{x \to x_0}f(x) = f(x_0)$ and $\lim \limits_{x \to x_0}g(x) = g(x_0)$", well we do not know that since we were not given that $f$ and $g$ are continuous functions. Even when $x_0\in D$ it may happen that $\lim \limits_{x \to x_0}f(x)\not = f(x_0)$ and or that $\lim \limits_{x \to x_0}g(x) \not= g(x_0)$. But as already discussed, the case when $x_0\in D$ is trivial, since then we may simply use that $f(x)\le g(x)$ is given for all $x\in D$. 
The above was only an example to help illustrate the statement of the problem, and to make it easier to motivate and appreciate the correct proof. 
It should be emphasized in any proof of the above problem, when considering the limits, that $x$ approaches $x_0$ but $x$ remains different of $x_0$ (whether or not $x_0$ belongs to $D$). To make this answer self-contained, here is a proof (which I tried to make look different than the two proofs already posted, to avoid repetition). Let 
$\lim \limits_{x \to x_0}f(x)=L$ and $\lim \limits_{x \to x_0}g(x)=M$. 
We need to prove that $L\le M$. It is enough to prove that for every $\varepsilon>0$ we have $L<M+\varepsilon$ (once this is done we may let $\varepsilon\to0$ to conclude $L\le M$). So fix $\varepsilon>0$. Then there is $\delta_1>0$ such that $L-\dfrac\varepsilon2<f(x)<L+\dfrac\varepsilon2$ whenever $x\in(x_0-\delta_1,x_0+\delta_1)$ with $x\not=x_0$. Similarly there is $\delta_2>0$ such that $M-\dfrac\varepsilon2<g(x)<M+\dfrac\varepsilon2$ whenever $x\in(x_0-\delta_2,x_0+\delta_2)$ with $x\not=x_0$. Let $\delta=\min\{\delta_1,\delta_2\}$ and using that $x_0$ is an accumulation point of $D$ pick $y\in D\cap(x_0-\delta,x_0+\delta)$ with $y\not=x_0$ (it is important that such a point $y\in D$ exists so we could compare $f(y)$ and $g(y)$). Then $L-\dfrac\varepsilon2<f(y)\le g(y)<M+\dfrac\varepsilon2$ hence $L<M+\varepsilon$ which completes the proof. 
A: The argument seems not substantial; I mean it looks to me like a restatement of what is to be proved.
We may argue by contradiction. Let $p:= \lim_{x \to x_{0}}f(x)$; let $q := \lim_{x \to x_{0}}g(x)$; let $p > q$. Then there is some $\delta > 0$ such that $0 < |x-x_{0}| < \delta$ implies
$$
|f(x) - p|, |g(x) - q| < \frac{p-q}{3};$$
but this implies that for $0 < |x-x_{0}| < \delta$ we have $f(x) > g(x)$, a contradiction.
A: Try this :
Put $h(x)=g(x)-f(x)$; It remains to show that 

If $h(x)\geq 0 \forall x $ then $\lim_{x\to x_0} h(x)\geq  0$

Proof: Suppose $\lim_{x\to x_0} h(x)=l$.Let $l<0$ then corresponding to $\epsilon=\dfrac{-l}{2}>0 \exists \delta >0 $  such that  $0<|x-x_0|<\delta \implies |h(x)-l|<\epsilon \implies \dfrac{3l}{2}<h(x)<\dfrac{l}{2}$
Since $l<0\implies h(x)<0 \forall x\in (x_0-\delta,x_0+\delta)\setminus \{x_0\}$ contradiction to $h(x)\geq 0 \forall x$
A: I am posting this just to add some diversity to the approaches taken to solve this problem. 
Recall that if the limit of a function exists, it is equal to the limit inferior and limit superior defined as follows : 
$$
\liminf_{x \to x_0} f(x) = \lim_{\delta \to 0} \left( \inf_{x \in ]x_0-\delta,x_0+\delta[ \backslash \{x_0\} } f(x) \right)
\limsup_{x \to x_0} f(x) = \lim_{\delta \to 0} \left( \sup_{x \in ]x_0-\delta,x_0+\delta[ \backslash \{x_0\} } f(x) \right).
$$
The function 
$$
\inf_{x \in ]x_0-\delta,x_0+\delta[ \backslash \{x_0\} } f(x)
$$
is monotone decreasing as a function of $\delta$, since if $\delta$ becomes smaller, the set over which we take the infimum contains less elements, thus the infimum can only grow, not decrease. Similarly, the function 
$$
\sup_{x \in ]x_0-\delta,x_0+\delta[ \backslash \{x_0\} } f(x)
$$
is a monotone increasing function of $\delta$. (Note that the fact that the infimum is monotone decreasing means that when $\delta$ becomes smaller, the infimum increases (and not decreases as the name "monotone decreasing" suggests) since we are thinking in the other direction). Assuming that the limit exists, both the infimum and supremum and continuous functions of $\delta$ at $\delta = 0$ and the limit of the infimum/supremum equals the limit of $f$ as $x \to x_0$. Also note that since both functions are monotone, we have
$$
\lim_{\delta \to 0} \left( \inf_{x \in ]x_0-\delta,x_0+\delta[ \backslash \{x_0\} } f(x) \right) = \sup_{\delta \to 0} \left( \inf_{x \in ]x_0-\delta,x_0+\delta[ \backslash \{x_0\} } f(x) \right), \qquad \lim_{\delta \to 0} \left( \sup_{x \in ]x_0-\delta,x_0+\delta[ \backslash \{x_0\} } f(x) \right) = \inf_{\delta \to 0} \left( \sup_{x \in ]x_0-\delta,x_0+\delta[ \backslash \{x_0\} } f(x) \right)
$$
Now that I have introduced the infimum and supremum, let's try to solve the exercise. The assumption $f(x) \le g(x)$ implies that for every $\delta > 0$ small enough so that this infimum is not $-\infty$,
$$
 \inf_{x \in ]x_0-\delta,x_0+\delta[ \backslash \{x_0\} } f(x) \le \inf_{x \in ]x_0-\delta,x_0+\delta[ \backslash \{x_0\} } g(x) 
$$
hence taking the supremum over all $\delta > 0$, we get
$$
\lim_{x \to x_0} f(x) = \sup_{\delta \to 0} \left( \inf_{x \in ]x_0-\delta,x_0+\delta[ \backslash \{x_0\} } f(x) \right) \le \sup_{\delta \to 0} \left( \inf_{x \in ]x_0-\delta,x_0+\delta[ \backslash \{x_0\} } g(x) \right) = \lim_{x \to x_0} g(x).
$$
Since the limit exists, we could have done the reasoning symmetrically using the limit superior instead. As you can see, the tools used are maybe a bit more advanced but the fact becomes much more trivial (which is why we create advanced tools in the first place!).
Hope that helps,
