# Definition of Slope of Tangent to a Curve Using Limits

I am restudying Calculus on my own, and I am a little bit stuck on the definition of the slope of a tangent line to a point on a curve. I understand the definition somewhat, but I got to wondering about why the current definition was chosen. Basically, I do not understand why we do not take the following limit as the definition: $$\text{slope} = \lim\limits_{Q\to P}\text{slope}_{\text{sec}} = \lim\limits_{y_1\to y_0}\frac{y_1-y_0}{x_1-x_0}$$ Instead we use the following: $$\text{slope} = \lim\limits_{Q\to P}\text{slope}_{\text{sec}} = \lim\limits_{x_1\to x_0}\frac{y_1-y_0}{x_1-x_0}$$ • How would you find the derivative of $f(x)=1$ then? The point is, if $f$ is a function, for any $x$, there is exactly one $y$ s.t. $f(x)=y$. – enedil Mar 7 '19 at 0:47
• @enedil Thanks for the comment. You have a good point. – SebastianLinde Mar 7 '19 at 15:35

Using a graph, you could do it the way you suggest relatively easily. However, more generally, if you are using a function definition instead, I believe the basic issue is that you would need to find the inverse of the function. As enedil commented, for $$x_1 \to x_0$$, you can determine $$y_1$$ for each $$x_1$$ easily using the function definition, i.e., $$f(x_1) = y_1$$. However, if you tried $$y_1 \to y_0$$ instead, to determine the corresponding values of $$x_1$$, you would need to use the function inverse, i.e., $$x_1 = f^{-1}(y_1)$$. However, not all functions have inverses and, even if they do, it requires more work to calculate them than necessary.
In summary, to have just one definition covering both using a graph and a function, it makes more sense to have $$x_1 \to x_0$$.