# Why don't we include $y$ approaches to $y_0$ as a limit? [closed]

How comes we only use $$x$$ approaches to $$x_0$$ when $$y$$ approaches $$y_0$$ is equally important. Why don't we include $$y$$ approaches $$y_0$$?

• Concretely, what if we had $y = \sin x$, and we wanted to find the derivative at $(0, 0)$? If we require $x \to 0$, then that gives us $y \to 0$. However, if we know that $y \to 0$, that doesn't mean we've got the right point. It could be the case that $x \to \pi$ and $y \to 0$, in which case the limit of the slope of the secant is $0$, which isn't the number we want. – Izaak van Dongen Jul 9 at 13:57
• Since y is a continuous function of x, and $y(x_0)= y_0$, as x approaches $x_0$, y necessarily approaches $y_0$. It is not necessary to say both. – user247327 Jul 9 at 13:57

Because $$y$$ depends on $$x$$... The points $$(x_0,y_0)$$ and $$(x_1,y_1)$$ belong to the graphic of a function $$f$$, so $$y_1=f(x_1)$$ and $$y_0=f(x_0)$$.