Proof that the derivative of a linear function is $0$.

Question

We have a function $f(x)=x$ which is defined and is continuous on the set $S$ of all real numbers.

The derivative at point $x$, $$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}h$$

Using the theorems of limits we can replace the above limit equation by:$$f'(x)=\frac{\lim_{h\to 0}{f(x+h)}-f(x)} {\lim_{h\to 0}h}$$

Since $f$ is continuous at all points $f(x+h) \to f(x)$ as $h\to 0$ so $$f'(x)=\frac{0} {\lim_{h\to 0}h}$$ so $f'(x)=0$.

What is wrong here ??? This is a continuation of my other question Proving the derivative is $0$ at the extremum and all derivatives are $0$.

Answer 1

that part is wrong: "Using the theorems of limits we can replace the above limit equation by:" you need to know both limits (of the denominator and enumarator) exist, are finite, denominator's limit $\not= 0$. in your case $\lim_{h \rightarrow 0} = 0$ so you can't do that.

Answer 2

One problem is that $\frac{0}{0}$, in general, is not well defined. The operation you have performed is invalid, as well, since the denominator limit is zero. To compute the derivative of the function, you would do something like this. Let $h>0$,

$$\frac{f(x+h)-f(x)}{h}=\frac{x+h-x}{h}=\frac{h}{h}=1.$$

The result holds for any $h$, so the limit must be $1$.