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I just recently learned the the definition of differentiable at a point in my multi-variable calculus class. The analogy between the multi-variable definition and that of the single variable uses the fact that the difference between the exact value of the function and the value of the tangent line approximation of the function goes to $0$ as $\Delta x$ goes to $0$. Later I forgot the definition, but I used this intuitive idea to come up with the following definition:

A function $f$ is said to be differentiable at a point $(x_0, f(x_0))$ if $$\lim_{\Delta x\to 0} f(x_0+\Delta x) -(f'(x_0)\Delta x + f(x_0))= 0.$$

This is not the same as the definition in my text which states:

A function $f$ is said to be differentiable at a point $(x_0, f(x_0))$ if there exists a continuous function $$\epsilon(\Delta x)=\frac{f(x_0+\Delta x)}{\Delta x} - f'(x_0)$$ such that $$f(x_0+ \Delta x)= f(x_0) + f'(x_0) \Delta x + \epsilon \Delta x$$ where $\epsilon \to 0$ as $\Delta x\to 0$.

I ask: What is the difference between these two definitions, if any? If they are the same, how do you get one definition from the other?

Edit:

The text that I got the second definition from was Calculus 7th Ed. by James Stewart. Chapter 2 Section 5 Equation 5.

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    $\begingroup$ The point is that the difference between $f(x_0+\Delta x) - f(x_0)$ and $f'(x_0)\Delta x$ must go to zero rapidly. Your definition just says it goes to zero... but of course it will, as long as $f$ is continuous at $x_0$, and regardless of what you assign $f'(x_0)$ to be, because it's the difference between two vanishing quantities. $\endgroup$ – mjqxxxx Apr 1 '16 at 20:12
  • $\begingroup$ @mjqxxxx It sounds like there is quite a difference between these two definitions. Is it possible that a function could satisfy the first definition and not the second? $\endgroup$ – Saud Molaib Apr 2 '16 at 0:12
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    $\begingroup$ Any smooth function will satisfy the first definition with any value of $f'(x_0)$ whatsoever; it doesn't pick out a unique value. $\endgroup$ – mjqxxxx Apr 3 '16 at 14:25
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    $\begingroup$ And non-differentiable functions like $f(x)=|x|$ (at $x=0$) will satisfy the first definition as well. $\endgroup$ – mjqxxxx Apr 3 '16 at 14:26
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If we just take limit from both sides of the second definition, we get the first one, because the whole limit is the sum of limits, and $\epsilon = o(\Delta x)$.

But actually the second one is more useful and tells more about the sense of differentiability, because it is about looking at the function as if it is "almost linear" somewhere.

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  • $\begingroup$ Your argument shows that the second definition implies the first, but does the first argument imply the second? $\endgroup$ – Saud Molaib Apr 2 '16 at 0:22
  • $\begingroup$ Straightforward -- no. $\endgroup$ – sooobus Apr 2 '16 at 20:41
  • $\begingroup$ Actually the definition of differentiation given in all coursebooks I have is the second $\endgroup$ – sooobus Apr 2 '16 at 20:42
  • $\begingroup$ Could you use a counterexample to prove your point? $\endgroup$ – Saud Molaib Apr 2 '16 at 21:06

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