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There are many topics with the derivative definition, but I couldn't find a precise answer to my doubts. In one of the formulation the derivative of a function in a given point $x_0$ is a number $a\in\mathbb{R}$ such as:

$$f(x_0+h)=f(x_0) + a\cdot h +r(x_0,h)$$

In this, the $f(x_0) + ah$ term is the "best" linear approximation of $f(x_0+h)$, and $r(x_0,h)$ is some reminder (or correction). Now, if we make $h \to 0$ we want the $r(x_0,h) \to 0$. However, such an approach will not provide the proper derivative definition, and we must make the following:

$$\lim_{h \to 0} \frac{r(x_0,h)}{h}=0$$

which means the $r(x_0,h)$ vanishes "faster" than $h$ when $h \to 0$. Is there are clear explanation why this entire fraction must vanish, rather than the reminder itself? With many thanks.

rk85
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  • If you are ok, you can accept the answer and set as solved. Thanks! – user Dec 29 '17 at 20:57
  • Partly. See my last post with an extra proof of the need for the $a$ to be unique, but without using the "standard" derivative definition. – rk85 Dec 29 '17 at 21:01
  • Please, if you are ok, you can accept the answer and set it as solved. Thanks! – user Feb 01 '18 at 22:47

3 Answers3

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As you noted it is necessary that:

$$\lim_{h \to 0} \frac{r(x_0,h)}{h}=0$$

to guarantee that the $r(x_0,h)$ vanishes "faster" than $h$ when $h \to 0$.

In this case we say that f(x) is differentiable at $x_0$ and $a=f'(x_0)$.

Indeed:

$$\lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h}=\lim_{h \to 0}\frac{a\cdot h+r(x_0,h)}{h}=\lim_{h \to 0} \left(a+\frac{r(x_0,h)}{h}\right)=a+\lim_{h \to 0} \frac{r(x_0,h)}{h}$$

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user
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  • OK, yes, I fully agree with your answer and the previous one. Perhaps I was not precise enough. The original definition with the limit of the difference quotient must the the result of "my" definition. In this way, what would be rationale behind the entire $r/h$ vanishing? In my trails to understand this definition I started with lines passing through the point $(x_0,f(x_0))$, and trying the find the "best" $a$ (slope) approximating the function for very small $h$". – rk85 Dec 28 '17 at 22:38
  • @Radek the definition $f(x_0+h)=f(x_0) + a\cdot h +r(x_0,h)$ with $r(x_0,h)$ such that $\lim_{h \to 0} \frac{r(x_0,h)}{h}=0$ is precisely the defintion of differential, that is the condition which guarantees that we can approximate the function by a linear expansion near $x_0$ with an error wich goes to zero faster than $h$. For functions of one variable this definition is equivalent to that of derivative and $a=f'(x_0)$ whereas for funtions of several variables the true general definition which guarantees the local linear approximation for a function is the differential concept. – user Dec 28 '17 at 22:51
  • Many thanks. Yes, I didn't use the word differential, but I fully agree. I think I understand now my doubts, and somewhat feel uneasy not understanding why $r$ must vanish faster than $h$ with $h\to 0$? – rk85 Dec 28 '17 at 22:58
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You write "Now, if we make $h \to 0$ we want the [that] $r(x_0,h) \to 0$."

Not quite. What you want is that $a$ is the derivative $f'(x_0)$, which is given by $$ \lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}h=\lim_{h \to 0}\frac{f(x_0) + a\cdot h +r(x_0,h)-f(x_0)}{h}=a+\lim_{h \to 0} \frac{r(x_0,h)}{h}. $$ So that this is really $a$ we need that $$ \lim_{h \to 0} \frac{r(x_0,h)}{h}=0. $$

John B
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I have been thinking a lot yesterday, and following the discussion with user Gimusi (thanks! see above) I was able to clarify my doubts and now pose the question with more precision. My aim is to define the derivative differently than the "standard" approach with the difference quotient. For that reason I consider a bunch of straight lines of the following form $y(x)=ax+b$. Let us now consider the line must pass through the point of the $(x_0,f(x_0))$ coordinates we get the following expression for every line as $y(x)=f(x_0)+a\cdot(x-x_0)$. This form already encodes the line will pass through the selected point and it can be controlled with the slope parameter $a$. Now, let us consider the change of the function $f$ can be expressed as the sum of the linear part and some remainder (as in the original question):

$$f(x_0+h)=f(x_0)+a\cdot h+r(x_0,h)$$

Now, the discussion concluded we need to have $r(x_0,h)$ vanishing faster than $h$ when $h \to 0$, that is:

$$\lim_{h \to 0} \frac{r(x_0,h)}{h}=0$$

and not just:

$$\lim_{h \to 0} r(x_0,h) = 0$$

My impression is that in the latter case will not provide the definision of the unique $a$ value. My feeling is that if we let just $r(x_0,h) \to 0$ this will be true for any $a$. Only by letting the $r/h \to 0$ we will be able to define a single $a$ out of many for different lined passing through the point. Can we somewhat prove such a statement? With many thanks in advance.

rk85
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  • The fact is that the tangent at $x_0$ satisfy the definition of linear approximation with $\lim_{h \to 0} \frac{r(x_0,h)}{h}=0$ and the tangent is unique, thus "a" is unique and correspond to the derivative in $x_0$. I've added a figure to my answer to better explain the point. I hope it can help! – user Dec 29 '17 at 21:37