![enter image description here](https://i.stack.imgur.com/KGFEr.png)

Start by assuming that function curves are made of an infinite amount of lines (i.e. look at the image above but instead of approximating it using a finite number of lines, use infinite lines). This intuition is supported by the arc length formula (integral of infinite "line approximations").

Let's call the point at the top ("the highest point") point $x$. Refer to the $n^{th}$ point before that as $x-n$ and the $n^{th}$ point after that as $x+n$. I want to calculate the derivative at point $x$. In other words I want to calculate the slope of the line connecting point $x$ and $x+1$.

I can start be drawing a secant line from point $x$ to $x+n$ and using the slope to approximate the derivative. As $n$ approaches $1$, you get better approximations of the derivative. As $n=1$ you exactly have the derivative. 

Now draw a secant line from point $x$ to $x-n$. As n approaches 1 you get infinitely better approximations of the slope of the line between $x$ and $x-1$, but not $x$ and $x+1$. No matter how close to n=1 you get you will never get the slope of the line that connects $x$ and $x+1$ (which is what the derivative is: the rate of change of the function between a point and the next infinitesimal point).

So why is the derivative the limit:

$$\lim \_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$$

instead of:

$$\lim \_{h \rightarrow 0^+} \frac{f(x+h) - f(x)}{h}$$

Since using the above intuition shows that the left hand limit doesn't give you the slope of the line between $x$ and the next infinitesimal point.