Proving an inequality of limits given an inequality of functions (Spivak's Calculus Chapter 5 Problem 12(a)) 

*(a) Suppose that $f(x) \le g(x)$ for all x. Prove that $\lim\limits_{x\to a}f(x)\le\lim\limits_{x\to a} g(x)$, provided that these limits exist.

The answer key, along with some other sources I saw online, provides a proof by contradiction. It starts by supposing that $\mathit l$ = $\lim_{x\to a}$ f(x) $\gt$ $\lim_{x\to a}$ g(x)=m. Then, it lets $\epsilon=\mathit l-m$. Then there is a $\delta\gt0$ such that if $0\lt$ $\vert x-a\vert$$\lt$$\delta$, then $\vert\mathit l-f(x)\vert$$\lt$$\epsilon$/2 and $\vert m-g(x)\vert\lt\epsilon$/2.
Thus for $0\lt\vert x-a\vert\lt\delta$ we have g(x) $\lt m+\epsilon/2=\mathit l-\epsilon/2\lt f(x)$, which contradicts f(x)$\le$ g(x).
I understand this answer, and I would have used it had it been the first thing I came up with. I instead put
$g(x)=f(x)+b$ for some variable $b\ge0$.  So, if for some $\delta \gt 0$,$0\lt\vert x-a\vert\lt\delta$, $\vert g(x)-m\vert\lt\epsilon$ $\implies$ $\vert f(x)+b-m\vert\lt\epsilon$ $\implies$ $\vert f(x)-(m - b)\vert\lt\epsilon$. So, the limit $\mathit l =m-b=$ $\lim_{x\to a}$ f(x) $\le$ $\lim_{x\to a} g(x)=m$.
Does this answer work as well?
 A: 
$g(x)=f(x)+b$ for some variable $b\ge0$.  So, if for some $\delta \gt 0$,$0\lt\vert x-a\vert\lt\delta$, $\vert g(x)-m\vert\lt\epsilon$ $\implies$ $\vert f(x)+b-m\vert\lt\epsilon$ $\implies$ $\vert f(x)-(m - b)\vert\lt\epsilon$. So, the limit $\mathit l =m-b=$ $\lim_{x\to a}$ f(x) $\le$ $\lim_{x\to a} g(x)=m$.

Note, you let $g(x)=f(x)+b$, but here $b$ depends on the point $x$, so $b=b(x)$, and you have to fix all the work afterwards.
Maybe an easier way to prove this problem is to define $h(x)=f(x)-g(x)$, in this case, we are given $h(x)\le 0,~\forall x\in \mathbb{R}$ and $\lim\limits_{x\to a}h(x)$ exists. We want to prove $\lim\limits_{x\to a}h(x)\le 0$
A: You can't assume that $g(x) = f(x) + b, b \ge 0$. That's one of the big mistakes even math students commit. They mainly use their feeling and arbitrary assumptions that help make things work out for them only. That's in the long run will be really bad for a math student. Look at $f(x) = x^2, g(x) = x^2 + \sin^2x$, then $f(x) \le g(x), \forall x$, but there is no $b \ge 0$ such that $g(x)= f(x)+b$,because if there were such $b$, then $b = \sin^2x$ for all $x$, and this is not possible as $b$ is a constant and $\sin^2x$ is not. That said, now you can show directly that if $L = \displaystyle \lim_{x \to a} f(x)$ and $K = \displaystyle \lim_{x \to a} g(x)$, then $L \le K$. So let $\epsilon > 0$ be given, there is a $\delta > 0$ such that if $0 < |x-a| < \delta$ then $|f(x) - L|< \epsilon, |g(x) - K| < \epsilon\implies -\epsilon < f(x) - L < \epsilon\implies L - \epsilon < f(x) \le g(x) < K + \epsilon\implies L - K < 2\epsilon$. Since this is true for any $\epsilon > 0$, it implies that $L - K \le 0$, and hence $L \le K$.
