Let $f(x) = x $ for $x \ge 0$ then $\lim_{x\to 0^+} f(x) = 1.$
Now, how could I exaplin it on a graph? I can't uderstand from the graph why the limit approaches $1$. It seems to me that when $x$ get closer and closer to $0$ it approaches $0$.
Edit:
The entire function is $f(x) = \begin{cases} x^2\textrm{ for $x \le 0$} \\ x \textrm{ for $x \ge 0$} \\ \end{cases}$
And I need to prove $f'(x)$ for $x = 0$
Then I've got $\frac{f(h) - f(0)}{h} = \begin{cases} \frac{h}{h} = 1 \textrm{ for $h > 0$} \\ h \textrm{ for $h<0$} \\ \end{cases}$
Then : $\lim_{x\to 0^+} \frac{f(h) - f(0)}{h} = 1$ and $\lim_{x\to 0^-} \frac{f(h) - f(0)}{h} = 0.$
It is actually a kind of impossible!!! When you have $\lim_{x\to 0^+} f(x) = 1.$, it means that for every $\eta>0$ there is a $\sigma>0$ such that $|x-0^+|<\sigma$, then $|f(x)-1|<\eta$. While in your case, we have $|f(x)-1|=|x-1|$ for $x\geq 0$, and so $|f(x)-1|=|x-1|\geq 1-\sigma$. So, I think there is no such a function.