# If $f(x)<g(x)\forall x \in \mathbb{R}$, then $\lim_{x\to 0}f(x) \leq \lim_{x\to 0}g(x)$?

Suppose $$f,g:\mathbb{R}\to\mathbb{R}$$ are functions for which $$f(x) for all $$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 $$f(x) for all $$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).

Am I on the right track?

• 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 $\geq 0$. – Ben W Feb 14 at 9:54
• Nothing to do with continuity. Use definition of limit to prove that this is true. – 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 $$h(x) for all $$x \in N,$$ a contradiction.