If $f(x) < g(x)$ for all $x$, $\lim f(x) \leq \lim g(x)$?

Question

Let $X$ be a metric space, and let $a$ be a limit point of $X$. Let $f,g:X\to\mathbb{R}$ be functions such that $f(x) < g(x), \forall x \in X$, and $\lim_{x \to a}f(x)$ and $\lim_{x\to a}g(x)$ exist. Prove that $$\lim_{x\to a}f(x) \leq \lim_{x\to a}g(x)$$ and give an example for which the limits are equal.

I'm stuck in epsilon-delta stuff (of which I have a pretty poor grasp anyway). How can I use the epsilon-delta definition of a limit to prove the above? Does continuity have anything to do with it?

Answer 1

Hint:

Setting $h(x)=g(x)-f(x)$, this amounts to showing that if $h(x)>0$ for all $x$, then $\;\lim\limits_{x\to a}h(x)\ge 0$.

Answer 2

To begin with, let's assume that $\displaystyle\lim_{x\rightarrow a} f(x) = L$, $\displaystyle\lim_{x\rightarrow a} g(x) = M$ and $f(x) < g(x)$.

We are going to prove that $L\leq M$.

According to the definition of limit, for every $\varepsilon > 0$, there are $\delta_{1} > 0$ and $\delta_{2} > 0$ such that \begin{align*} \begin{cases} 0 < |x - a| < \delta_{1} \Rightarrow |f(x) - L| < \varepsilon\\\\ 0 < |x - a| < \delta_{2} \Rightarrow |g(x) - M| < \varepsilon \end{cases} \Longrightarrow L - \varepsilon < f(x) < g(x) < M + \varepsilon \end{align*}

If we assume that $L > M$, for $\displaystyle\varepsilon = \frac{L - M}{3}$, there is a $\delta = \min\{\delta_{1},\delta_{2}\}$ such that \begin{align*} M - L + 2\epsilon = M - L + \frac{2(L-M)}{3} = \frac{M - L}{3} > 0 \Longrightarrow M > L \end{align*} which contradicts our assumption. Thus $M\geq L$, and we are done.