Vtyjkyuk

1. If a function $f(x)$ is continuous and increasing at point $x=a,$ then there is a nbhd $(x-delta,x+delta),delta>0$ where the function is also increasing.


2. if $f' (x_0)$ is positive, then for $x$ nearby but smaller than
   $x_0$ the values $f(x)$ will be less than $f(x_0)$, but for $x$
   nearby but larger than $x_0$, the values of $f(x)$ will be larger
   than $f(x_0)$. This says something like $f$ is an increasing
   function near $x_0$, but not quite.

The two statements are not the same. One assumes differentibity and the other does not.

Part 1.-

Consider $f(x)=xleft(sin frac1x +2right), x>0, f(0)=0.$ It is clear that for $x>0$ it is $f(x)>0.$ But $$f'(x)=sinfrac1x+2-frac1xcosfrac1x.$$ So $f$ is increasing at $0$ but not increasing in $(0,delta).$ (Of course, you can modify the function to work on $(-delta,delta).$

Edit

An example can be found in the nice book (see Problem 1.13)
https://stemtec.aut.ac.nz/__data/assets/pdf_file/0003/57639/Counterexamples-in-Calculus-MAA-e-book.pdf

Part 2.-

If $f'(x_0)=lim_xto x_0dfracf(x)-f(x_0)x-x_0>0$ we have that there exists $delta>0$ such that $xin (x_0-delta,x_0)implies f(x)<f(x_0)$ and $xin (x_0,x_0+delta)implies f(x)>f(x_0).$ Thus $f$ is increasing at $x_0.$

Conclusion

Both statements are not the same. The first one is false while the second one holds.