# Thread: Question of limits of functions

1. ## Question of limits of functions

Suppose that f and g are real valued functions defined on some interval (a,b) containing the point $\displaystyle x_0$. Suppose also that $\displaystyle \lim_{x \to x_0}f(x) =A$ and $\displaystyle \lim_{x \to x_0}f(x) =B$. Prove that if $\displaystyle f(x)<g(x)$ for all $\displaystyle x \in (x_0 - \phi,x_0+\phi)$ (for some $\displaystyle \phi >0$). Then $\displaystyle A \leq B$. 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 $\displaystyle x_0$ with $\displaystyle \epsilon, \delta$ 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!

2. Originally Posted by worc3247
Suppose that f and g are real valued functions defined on some interval (a,b) containing the point $\displaystyle x_0$. Suppose also that $\displaystyle \lim_{x \to x_0}f(x) =A$ and $\displaystyle \lim_{x \to x_0}f(x) =B$. Prove that if $\displaystyle f(x)<g(x)$ for all $\displaystyle x \in (x_0 - \phi,x_0+\phi)$ (for some $\displaystyle \phi >0$). Then $\displaystyle A \leq B$. In this case is it always true that A<B.
To prove that $\displaystyle A\le B$ assume that if $\displaystyle B<A$ then let$\displaystyle \epsilon=\frac{A-B}{4}}>0$.

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

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

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

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

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

Suggestion draw a picture.

6. 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.