# Can $\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}$ be written as $\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$ when $h=x-x_0$?

Say that we have a function $f(x)=x$, and two distinct points $P(x_0,f(x_0))$ and $Q(x,f(x))$ on the curve $y=f(x)$; if we let Q approach P, then the slope of the secant through Q and P will approach a limit, that is the slope of the tangent line at P: $$m_{tan}= \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0} . . . . . (1)$$ I have read that if we let $h$ denote the difference $h=x-x_0$, then the statement $x\to x_0$ is equivalent to the statement $h\to 0$, so we can rewrite $(1)$ in terms of $x_0$ and $h$ as:$$m_{tan}= \lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h} . . . . . (2)$$ Okay, if someone said that since $h$ denote the difference $h=x-x_0$, then the statement $h\to 0$ is equivalent to the statement $x_0\to x$, so we can rewrite $(2)$ as:$$m_{tan}=\lim_{x_0\to x}\frac{f(x)-f(x_0)}{x-x_0} . . . . . (3)$$ BUT, this is totally wrong since $(1)$ and $(3)$ are totally different limits that give slops of tangents of two different points(namely, $P$ and $Q$).

Can $(1)$ be written as $(2)$? If yes, why can't $(2)$ be written as $(3)$?

Yes of course those are two equivalent ways to write the limits by a change of variables.

Note also that in the definition $x\to x_0$ the value $x_0$ is fixed and $x$ is approching that value at the point $P$.

Therefore, to summarize, starting from

$$\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}$$

by the change of variable $x=x_0+h \iff x-x_0=h$ we have that $x\to x_0 \iff h\to 0$ and

$$\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$$

• Maybe you didn't read to the end. Won't $h→0$ not only when $x→x_0$ but also when $x_0→x$? If yes then $\lim_{x→x_0}\frac{f(x)−f(x_0)}{x−x_0}=\lim_{h→0}\frac{f(x_0+h)−f(x0)}{h}=\lim_{x_0→x}\frac{f(x)−f(x0)}{x−x_0}$ Commented Sep 16, 2018 at 22:00
• but the first limit here is the slope of the tangent at P, and the last is the slope at Q! Commented Sep 16, 2018 at 22:03
• @مُحَمَّدْ Yes I've read that point and as specified in the definition we are considering $x_0$ fixed.
– user
Commented Sep 16, 2018 at 22:03
• @مُحَمَّدْ Both are the slope of the tangent at $x=x_0$ that is at $P(x_0,f(x_0))$.
– user
Commented Sep 16, 2018 at 22:05
• So $h\to 0$ means $x \to x_0$ but not $x_0 \to x$ because we are considering $P$ fixed? Commented Sep 16, 2018 at 22:10

One way to see what is really going on with (1) (2) and (3) is looking at concrete examples. You let $f(x)=x$ at the beginning of your post but never use it again.

• In (1), $$\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0} = \lim_{x\to x_0}\frac{x-x_0}{x-x_0}=1.$$
• In (2), $$\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}=\lim_{h\to 0}\frac{(x_0+h)-x_0}{h}=1.$$

Let's consider another example: $f(x)=x^2.$

• In (1), $$\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0} = \lim_{x\to x_0}\frac{x^2-x_0^2}{x-x_0} = \lim_{x\to x_0} (x+x_0)=2x_0.$$
• In (2), $$\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}=\lim_{h\to 0}\frac{(x_0+h)^2-x_0^2}{h} =\lim_{h\to 0}\frac{(x_0+h+x_0)h}{h}=\lim_{h\to 0}2x_0+h=2x_0.$$

the statement $h→0$ is equivalent to the statement $x_0→x$

No! It is not that they are equivalent or not, but that "$h→0$" and "$x_0→x$" are not mathematical statements at all.

Can (1) be written as (2)?

Both (1) and (2) have the same value whenever exist. Either one of them, when exists, defines the derivative of $f$ at $x_0$.

If yes, why can't (2) be written as (3)?

In the case when the notation $x_0$ is some given fixed number, the expression $$\lim_{x_0\to x}\frac{f(x)-f(x_0))}{x-x_0}$$ is meaningless. It is something like $$\lim_{1\to x}\frac{f(x)-f(1)}{x-1},$$ which is meaningless. If you are considering the following two expressions: $$\lim_{x\to y}\frac{f(x)-f(y)}{x-y},\quad \lim_{y\to x}\frac{f(x)-f(y)}{x-y}$$ they are in general two different things. Again, using the example $f(x)=x^2$, you can see that $$\lim_{x\to y}\frac{f(x)-f(y)}{x-y}=2y,$$ but $$\lim_{y\to x}\frac{f(x)-f(y)}{x-y}=2x.$$

• Okay, thank you, but I still got some problems. Can't I write $(1)$ as: $\lim_{h\to0}\frac{f(x)-f(x-h)}{h}$? If I solved this for $f(x)=x^2$, I will get $2x$, which is the value of $(3)$ for $f(x)=x^2$ assuming that $x_0$ is a variable and $x$ is a constant! Commented Sep 18, 2018 at 2:24
• @مُحَمَّدْ I think one should stick to one set of notations to avoid all possible confusion. Mathematically, there is no difference among the expressions: $f(x), f(z), f(y), f(x_0), f(\text{whatever letter you would like to put here})$, etc. However, in a given piece of argument, once one defines/claims a set of notation, one should not change the meaning of it anymore. For instance, given that in your example, $x_0$ is a constant, one should never use the symbol $x_0$ as a "variable" anymore. (cont.)
– user587192
Commented Sep 18, 2018 at 14:16
• (cont.) There are many equivalent formulations of the derivative $f'(x)$: for instance $f'(x)= \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$, $f'(x)=\lim_{y\to x}\frac{f(y)-f(x)}{y-x}$, $f'(x)=\lim_{h\to 0}\frac{f(x-h)-f(x)}{-h}=\lim_{h\to 0}\frac{f(x)-f(x-h)}{h}$. Without a specific goal in mind, I don't find algebraic manipulation of theses expressions very interesting.
– user587192
Commented Sep 18, 2018 at 14:16
• if I defined $x_0=c, \text{c is a constant}$, but also $x_0=h+x, \text{h and x are variables}$. Is $x_0$ a constant or a variable? That is what I did to turn $(2)$ into the limit in my comment. Commented Sep 20, 2018 at 1:56

If you keep the point $p$ fixed and move the point $q$ then you find the slope at $p$

On the other hand if you keep $q$ fixed and move $p$ towards $q$ you get slope at point $q$

Usually they keep $x_o$ fixed and let $x$ approach $x_0$

• Fine, but the book says that we can rewrite $(1)$ as $(2)$[since $x\to x_0$ is equivalent to $h\to 0$]. Hence, can we rewrite $(2)$ as $\lim_{x\to x_0}\frac{f(x_0)-f(x)}{x_0-x}$[since $h\to 0$ is equivalent to $x_0\to x$]!!? but this limit and $(1)$ are totally different limits that will give slopes of tangents at different points! Commented Sep 16, 2018 at 22:25
• $\lim_{x\to x_0}\frac{f(x_0)-f(x)}{x_0-x}= \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0} = f'(x_0)$ You are multiplying top and bottom by a negative one. That does not change the fraction. Commented Sep 16, 2018 at 22:31
• Sorry. I meant to write the limit in my comment as $\lim_{x_0\to x}\frac{f(x)-f(x_0)}{x-x_0}$ Commented Sep 16, 2018 at 22:41
• That is $f'(x)$ which is different from $f'(x_0)$ as you have said before. Commented Sep 16, 2018 at 22:45