Does the given statements means the same?

Question

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.

Answer 1

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

Answer 2

Part 1.-

Consider $f(x)=x\left(\sin \frac1x +2\right), x>0, f(0)=0.$ It is clear that for $x>0$ it is $f(x)>0.$ But $$f'(x)=\sin\frac1x+2-\frac1x\cos\frac1x.$$ 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 $f'(x_0)=\lim_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}>0$ we have that there exists $\delta>0$ such that $x\in (x_0-\delta,x_0)\implies f(x)<f(x_0)$ and $x\in (x_0,x_0+\delta)\implies f(x)>f(x_0).$ Thus $f$ is increasing at $x_0.$

Conclusion

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