Order Limit Theorem for Functional Limits Let $f$ and $g$ be functions defined on a set $A$ and assume $f(x) \le g(x)$ for all $x \in A$. Further, let $c$ be a limit point of $A$ and assume that the limits for $f$ and $g$ exist at $c$. Show that $$\lim_{x\to c} f(x) \le \lim_{x\to c} g(x)$$.
I know that it makes sense I'm just not sure how to prove it
This is for interpreting the question as having $f,g$ functions defined on a set $A$ of reals including all $x$ near but not (necessarily) equal to $c.$ Let the limits as $x \to c$ of $f,g$ be respectively denoted $F,G,$ and recall one is assuming $f(x) \le g(x)$ for all $x \in A.$ 

We wish to show the claim that $F \le G.$ Setting things up for a contradiction proof, we assume to the contrary that $F&gt;G.$ Now define $H=(F+G)/2,$ and note since $f(x) \to F$ there is $\delta_1&gt;0$ for which $f(x)&gt;H$ as long as $|x-c|&lt;\delta_1.$ [Apply here that $f \to F &gt;H.$] Similarly there is $\delta_2$ for which $g(x)&lt;H$ as long as $|x-c|&lt;\delta_2.$ Then if $\delta = \min(\delta_1,\delta_2)$ and $|x-c|&lt;\delta,$ we have both $f(x)&gt;H$ and $g(x)&lt;H.$ Combining these gives $g(x)&lt;H&lt;f(x),$ i.e. $g(x)&lt;f(x),$ going against our assumption that $f(x) \le g(x)$ for all $x \in A.$
This is for interpreting the question as having $f,g$ functions defined on a set $A$ of reals including all $x$ near but not (necessarily) equal to $c.$ Let the limits as $x \to c$ of $f,g$ be respectively denoted $F,G,$ and recall one is assuming $f(x) \le g(x)$ for all $x \in A.$ 

We wish to show the claim that $F \le G.$ Setting things up for a contradiction proof, we assume to the contrary that $F&gt;G.$ Now define $H=(F+G)/2,$ and note since $f(x) \to F$ there is $\delta_1&gt;0$ for which $f(x)&gt;H$ as long as $|x-c|&lt;\delta_1.$ [Apply here that $f \to F &gt;H.$] Similarly there is $\delta_2$ for which $g(x)&lt;H$ as long as $|x-c|&lt;\delta_2.$ Then if $\delta = \min(\delta_1,\delta_2)$ and $|x-c|&lt;\delta,$ we have both $f(x)&gt;H$ and $g(x)&lt;H.$ Combining these gives $g(x)&lt;H&lt;f(x),$ i.e. $g(x)&lt;f(x),$ going against our assumption that $f(x) \le g(x)$ for all $x \in A.$