Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

4 Answers

up vote 6 down vote accepted

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.

share|improve this answer
    
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
add comment

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?

share|improve this answer
add comment

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 $a=0$.

share|improve this answer
add comment

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 $A=B$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.