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$$\frac{dy}{dx} = \lim_{\partial x\to 0} \frac{\partial y}{\partial x}$$

This is the general result to find the gradient of a line at any point on a curve.

So, my question is, why is it only $\lim_{\partial x\to 0}$ and not $\lim_{\partial y\to 0}$ too?

Because from what I understand. we're trying to make the secant line be as close as it can to the tangent line, so close that the difference between the gradients of both lines can be ignored. Hence we can approximately assume the gradient of the tangent line!

However, if we include both $\lim_{\partial x\to 0}$ and $\lim_{\partial y\to 0}$, wouldn't the gradient always be $\approx 1$?

I know there must be something wrong in my understanding of the whole concept, or perhaps I grasped the wrong idea of what Differentiation really is. I've tried watching Khan Academy videos and other related ones on Youtube too, but I feel like the more I watch them the more I'm confusing myself.

I've also tried to understand the concept from my textbook, but it just doesn't work. I tried completing the exercises my teacher gave, in hope that I might understand something.

Thank you! Hopefully you'll enlighten me.

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    $\begingroup$ $\lim_{x\to0}\frac{3x}{x}=3$ and not $1$. $\endgroup$ – Michael Hoppe Jul 30 '17 at 11:16
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First off, I think the notation you want to use is $$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$$

$\partial$ is usually used for partial derivatives, which doesn't make much sense in the way you're using it.

So, my question is, why is it only $\lim_{\partial x\to 0}$ and not $\lim_{\partial y\to 0}$ too?

$\Delta y$ does indeed tend to $0$. However, this isn't assumed, since if $\Delta x \to 0$, $\Delta y \to 0$ as well, since continuity is weaker than differentiability. Or, algebraically, if you consider

$$\frac{f(x)}{g(x)}$$

and the denominator $g(x)$ tends to $0$ as $x \to c$, then the numerator $f(x)$ must also tend to $0$ as $x \to c$ for the ratio to have a finite limit.

However, if we include both $\lim_{\partial x\to 0}$ and $\lim_{\partial y\to 0}$, wouldn't the gradient always be $\approx 1$?

No, because $$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$$ is a $\frac{0}{0}$ indeterminate. It can be any number, including $± \infty$.

Geometrically, this is because tangent lines to curves can be arbitrarily steep.

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