Question of limits of functions

**worc3247** · March 16th 2011, 01:46 PM

Suppose that f and g are real valued functions defined on some interval (a,b) containing the point $x_0$ . Suppose also that $\lim_{x \to x_0}f(x) =A$ and $\lim_{x \to x_0}f(x) =B$ . Prove that if $f(x)<g(x)$ for all $x \in (x_0 - \phi,x_0+\phi)$ (for some $\phi >0$ ). Then $A \leq B$ . In this case is it always true that A<B.

For the first part my proof so far consists of a contradiction. I have assumed A>B and then proceded to write out the definitions of limit tending to $x_0$ with $\epsilon, \delta$ and so on. However I can't seem to find a contradiction and was wondering if someone could give me the first few lines I need.
I also have no idea for the second part. Thanks!

**Plato** · March 16th 2011, 01:56 PM

Originally Posted by worc3247

Suppose that f and g are real valued functions defined on some interval (a,b) containing the point $x_0$ . Suppose also that $\lim_{x \to x_0}f(x) =A$ and $\lim_{x \to x_0}f(x) =B$ . Prove that if $f(x)<g(x)$ for all $x \in (x_0 - \phi,x_0+\phi)$ (for some $\phi >0$ ). Then $A \leq B$ . In this case is it always true that A<B.

To prove that $A\le B$ assume that if $B<A$ then let $\epsilon=\frac{A-B}{4}}>0$ .
A contradiction quickly follows.

**worc3247** · March 17th 2011, 05:44 AM

Sorry, I still can't see what contradiction to get. I get that:
$0<|f(x)-A|+|g(x)-B|<\frac{A-B}{2}<$ and then can't see where to go from there.

**worc3247** · March 17th 2011, 06:12 AM

Can you argue like this:
$0<|f(x)-A|+|g(x)-B|<\frac{A-B}{2}<\lim_{x\to\{x_0}}f(x) - \lim_{x\to\{x_0}}g(x)=\lim_{x\to\{x_0}}{(f(x)-g(x))}<0$ which is a contradiction?

**Plato** · March 17th 2011, 06:32 AM

Here is the idea.
Let $t \approx x_0$ stand for $|t-x_0|<\delta$ for some $\delta>0$ .

From the given we know that $t \approx x_0$ means that
$f(t) \approx A$ and $g(t) \approx B$ .

But if $B<A$ we are given that $f(t)\le g(t)$ there is a contradiction in there.

Suggestion draw a picture.

**LoblawsLawBlog** · March 17th 2011, 11:18 AM

Originally Posted by worc3247

I also have no idea for the second part. Thanks!

There might be an even simpler example, but since the limit has to be taken at an interior point of an interval I could only think of doing something like f(x)=0, g(x)=|x| if x is nonzero and g(0)=1. These are both defined on (-1,1) and g>f there. Now take limits as you approach 0.

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