$\displaystyle{\lim_{\Delta x \to 0} f(x+\Delta x) = f(x)}$

Wikipedia states that:

$\displaystyle{\lim_{\Delta x \to 0} f(x+\Delta x) = f(x)}$

Link: https://en.wikipedia.org/wiki/Product_rule#Proofs (In Proof by factoring from first principles)

But then:

$\displaystyle{\lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}}$ = $\displaystyle{\lim_{\Delta x \to 0} \frac{f(x) - f(x)}{\Delta x}}$ = 0.

Is the statement wrong or my conclusion ?

Solution 1:

Your mistake is in replacing $f(x+\Delta x)$ with $f(x)$ in the limit calculation. A simpler (non-)example:

It is true that $\lim_{x\to0}\frac{x}{x}=\lim_{x\to 0}1=1$. But if you only replace the numerator with its (correct) limit you get the incorrect $\lim_{x\to 0}\frac{0}{x}=\lim_{x\to 0}0=0$. This shows you can't pick a subexpression and substitute it for its limit.

This is different from for example evaluating $\lim_{x\to 0} \frac{1+x}{2+x}=\frac{1}{2}$ by replacing both the numerator and the denominator by their respective limits. That would be correct and justified because of the theorem that if $\lim_{x\to a} f(x)=L$ and $\lim_{x\to a} g(x)=M$ then $\lim_{x\to a} \frac{f(x)}{g(x)}=\frac{L}{M}$ (given $M\ne 0$). Notice that when we are using this theorem (and similar theorems for the other operations of arithmetic) we are left with a number as a result and not with another limit.