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When I used to prove in high school that $f'(x) = (f(x) = c)' = 0$ I did the following:

$$ \lim_{h \rightarrow \infty} \frac{f(x+h)-f(x)}{h} = \frac{c-c}{h} = 0 $$

but now the notation in my course at the university is a little bit different so I am a little confused. Here is the definition in my book:

Let $A \subseteq \mathbb{R}$, $f: A \rightarrow \mathbb{R}$ be a function and a point in $A$. We say that f is differentiable in $a$ if $a$ is an inner point and

$$ f'(a) = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a} $$ So for proving that $f'(x) = 0$ for $f(x) = c$ I have $$ f'(a) = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a} = \frac{c-c}{x-a} = 0 $$ and we are done. Is this just it? But I am little unsure why $f(a) = c$. Is it because $f : A \rightarrow \mathbb{R}$ is a function defined as $f(x) = c$ where the point $a \in A$ so $f(a) = c$? Thanks.

Thanks.

Mathias
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1 Answers1

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Important to realize that the only 2 real differences between your high school definition and the university definition is that the high school definition assumes the derivative always exists at all points and gives a formula for solving it as a function for all values of $x$ whereas the university definition only assumes you are only assuming it exists for a specific point. The high school definition also assumes the domain of the function is $\mathbb R$.

to show the two definitions are essentially the same we can consider that if $f$ has $\mathbb R$ as it's domain. And we want to find $f'(x)$ for any particular $x$ and not just a specific $a$... then we treat $x$ as fixed (it'll play the university role as "$a$") and we'll consider the real values that "approach" $x$-- we'll need a variable name; why not $y$?-- so but the university definition

$f: \mathbb R \to \mathbb R$ then $f'(x)$ where we consider $x$ as fixed but arbitrary and hence $f'(x) $ is a formula for the derivatives in general, the university definition gives

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

Now the assumption is that $y\ne x$.... so let $h = y-x$ then $y = x+h$ and we have

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

$\lim_{h\to x}\frac {f(x+h) - f(x)}h$.

Same thing.

......

Why am I pointing that out?

Well, I guess because you are having a disconnect between notation and concept. The notation changes because you have to be more precise and careful. But that doesn't mean things change.

If $f:A\to \mathbb R$ so that for all $f(x)= c$ for all $x \in A$ then for any $a\in A$ then.... by definition $f(a) = c$. Because that is how the function is defined. So for and $x\ne a$ but $x\in A$ we will have $f(x) = c$. And for $a$ we will have $f(a) = c$. If we had $b,\omega,d,e \in A$ we also have $f(b)=f(\omega) = f(d) =f(e) = c$.

And $f'(a) = \lim_{x\to a} \frac {f(x) -f(a)}{x-a}=\lim_{x\to a}\frac{c-c}{x-a}=\lim\frac 0{x-a}= 0$.

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BTW there are a few errors in your presentation I just have to correct:

$\lim_{h \rightarrow \color{red}0} \frac{f(x+h)-f(x)}{h} = \color{red}{\lim_{h\to 0}}\frac{c-c}{h}\color{red}{=\lim_{h\to 0}\frac{0}{h}} = 0$

First of all you had what I assume was a typo. The limit goes to $0$.

Second. Picky of me but important. $h$ is variable and not set. So you can't claim anything is equal to $\frac {c-c}h$. You are still talking about the limit as $h$ goes to $0$. Your aren't evaluating for any actual $h$.

Similarly in the university definition:

$f'(a) = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a} = \color{red}{\lim_{x\to a}}\frac{c-c}{x-a} = 0$

fleablood
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