# If $f(x)<g(x)$ prove that $\lim f(x)<\lim g(x)$

I have this question:

Let $f(x)→A$ and $g(x)→B$ as $x→x_0$. Prove that if $f(x) < g(x)$ for all $x∈(x_0−η, x_0+η)$ (for some $η > 0$) then $A\leq B$. In this case is it always true that $A < B$?

I've tried playing around with the definition for limits but I'm not getting anywhere. Can someone give me a hint on where to start?

-

HINT: Suppose that $A>B$, and let $\epsilon=\frac12(A-B)>0$. Show that there is a $\delta>0$ such that $f(x)>A-\epsilon=B+\epsilon>g(x)$ for all $x\in(x_0-\delta,x_0+\delta)$; this contradicts the hypothesis that $f(x)<g(x)$ on an open interval around $x_0$. (Why?)

It’s easy to find examples in which $A=B$. You can do it with $f(x)=0$, in fact.

-
Thanks, I've got it now. – Joe Jan 27 '13 at 10:06
@Joe: Excellent; you’re welcome. – Brian M. Scott Jan 27 '13 at 10:17

To show it is not always the case that $A<B$, you can come up with an example to show it is possible for $A = B$. So if we let $f(x) = (\frac{1}{x})^2$ and $g(x) = (\frac{1}{x})^4$, we know $f(x) < g(x)$ $\forall x \in (-1,1)$. And we know $\lim_{x \to 0} f(x) = \lim_{x \to 0} g(x) = \infty$ so it is possible for $A=B$.

Note the interval I gave you above is of the form $(x_0-η,x_0+η)$, which is centered about $x_0$ (distinguishing this from other answers).

-

This is a standard problem using $\epsilon-\delta$ definition. You can assume $B=0$ by replacing $f'=f-g,g'=g-g=0$. Then you are in the situation that $f'<0$ for all $x$ in $x_{0}$'s small enough neighhorhood, then $\lim_{x\rightarrow x_{0}}f'<0$ as well. It is not difficult to construct a proof based on contradiction. For example consider a translation of above statement into $f'<\epsilon$ in the neighorhood for any given $\epsilon>0$. Then if $\lim_{x\rightarrow x_{0}}f'>0$, what does this tell you?

-

Claim: Let $k\colon\mathbb R \rightarrow\mathbb R$ be a function such that $\forall x \in\mathbb R\colon k(x)>0$ and let for $x=a$ the $\lim_{x \rightarrow a} k(x)$ exists. Then this limit is greater than or equal to $0$.

Proof: Let the limit be $l$ and assume that $l<0$. From the definition of limits we know that $\forall \epsilon>0 \exists \delta(\epsilon)>0$ such that whenever $0<|x-a|<\delta(\epsilon)$ then $|k(x)-l|<\epsilon$.

Now taking $\epsilon=-l/2$ and removing the modulus from the inequality we have $3l/2<k(x)<l/2$ for all $x$ such that $0<|x-a|<\delta(-l/2)$. This contradicts the assumption and proves the claim.

Now taking $k(x)=g(x)-f(x)$ and using algebra of limits we get the required result. Both the limits can be equal.

E.g., take the domain to $(0,\infty)$, $g(x)=3^x$, $f(x)=2^x$ and $a=0$.

-

Quite obvious from the definition of continuity. By continuity,$\forall \epsilon$ there exist $\delta$ s.t $x\in B_{\delta}(x_0) \implies|f(x)-A|<\epsilon$ and $|g(x)-B|<\epsilon$ and so we have $B-\epsilon<g(x) \ \text{and} f(x)<\epsilon +A$ so $B<2\epsilon +A \ \forall \epsilon >0$. Since $\epsilon$ is arbitrary so we conclude that $B<A$. But i am not sure what is missing, i am not sure why i couldn't prove that $A=B$

-