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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 $y_0$?

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  • $\begingroup$ 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. $\endgroup$ – Izaak van Dongen Jul 9 at 13:57
  • $\begingroup$ 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. $\endgroup$ – user247327 Jul 9 at 13:57
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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 $y_0=f(x_0)$.

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  • $\begingroup$ Thank you very much $\endgroup$ – Gonja Jul 9 at 13:55

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