$\epsilon$-$\delta$ proof I found an interesting problem in my textbook, it asks to prove the following statement:

If $f'(x_{0}) >0$, then there is a $\delta >0$ such that $f(x)<f(x_{0})$ if $x_{0}- \delta < x <x_{0}$, and $f(x) > f(x_{0})$ if $x_{0} <x < x_{0} + \delta$.

How would one prove this from the definitions?
 A: Since $f'(x_0) > 0$, we can find $\delta > 0$ so that:
$$
|x - x_0| < \delta \implies \frac{f(x) - f(x_0)}{x - x_0} > 0
$$
(why?)
Now, the quotient above is positive iff the numerator and denominator have the same sign. Split this into two cases and handle each case separately to get the desired result.
A: Let $m=f'(x_0)>0$.  Suppose by way of contradiction that $f(x)>f(x_0)$ did not hold in some interval $(x_0, x_0+\delta)$.  Then there would be some sequence $x_1, x_2, \ldots$ such that $x_i>x_0$, $\lim_{i\rightarrow \infty}x_i=x_0$, and $f(x_i)<f(x_0)$.  But then this contradicts $$\lim_{x\rightarrow x_0+}\frac{f(x)-f(x_0)}{x-x_0}=m>0$$
A: If $f(x_0)>0$ then for all sufficiently small $\epsilon>0$ there is $\delta>0$, whit $x<x_0$, such that 
\begin{align}
0<|x-x_0|<\delta
 \implies
&
\left|\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0) \right|<\epsilon 
\\
-\delta <x-x_0<\delta
\implies
&
0<f'(x_0) -\epsilon <\frac{f(x)-f(x_0)}{x-x_0} <+f'(x_0) +\epsilon
\\
x_0-\delta <x<x_0+\delta
\implies
&
0<f'(x_0)(x-x_0)-\epsilon (x-x_0)+f(x_0)< f(x) 
\end{align}
