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 $h=x-x_0$?

Question

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 $(3)$?

Answer 1

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

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

Answer 2

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$

Answer 3

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

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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 $$ \lim_{y\to x}\frac{f(x)-f(y)}{x-y}=2x. $$