Suppose $f,g:\mathbb{R}\to\mathbb{R}$ are functions for which $f(x)<g(x)$ for all $x \in \mathbb{R}$, and both $f$ and $g$ have limits as $x\to 0$. Then $\lim_{x\to 0}f(x) \leq \lim_{x\to 0}g(x)$.

I'm trying to decide whether or not this statement is true. It seems clear that if both $f$ and $g$ are continuous on the reals then $\lim_{x\to 0}f(x)$ is necessarily smaller than that of $g(x)$ given that $f(x)<g(x)$ for all $x \in \mathbb{R}$, (including $x=0$), using the definition of continuity of a function.

So we require functions with discontinuity at $x=0$ to find a counterexample to the statement (I think). That is, we need $f$ and $g$ such that their respective limits at $0$ do not equal their outputs at $0$, and that $f(0)<g(0)$.

Am I on the right track?

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    $\begingroup$ The statement is true. Can you visualize it? Note that this is equivalent to saying that if $h(x)=g(x)-f(x)>0$ and a limit exists then it is $\geq 0$. $\endgroup$ – Ben W Feb 14 at 9:54
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    $\begingroup$ Nothing to do with continuity. Use definition of limit to prove that this is true. $\endgroup$ – Kavi Rama Murthy Feb 14 at 9:57

Let $h=g-f$, then $h>0$ on $\mathbb R$ and $L:=\lim_{x\to 0}h(x)$ exists. You have to show that $L \ge 0.$

Assume to the contrary, that $L<0.$ Then there is a neighborhood $N$ of $0$ such that $h(x) <L/2 <0$ for all $x \in N,$ a contradiction.


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