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As you know by the eplsilon-delta definition of the limit,

$$0<{|x-c|}<δ$$

So $x$ cannot be $c$ in

$$f'(x)= \lim_{x \rightarrow c} \frac{f(x)-f(c)}{x-c} \dots (*)$$

So $\Delta x$ also cannot be $0$.

Then how to plug in zero for $\Delta x$ when evaluating $(*)$ with $\Delta x $ difference?

Like:

f(x)=x^2,f'(1)
lim Δx->0 f(1+Δx)-f(1)/Δx
lim Δx->0 (1+Δx)^2-1/Δx
lim Δx->0 Δx^2+2Δx/Δx
lim Δx->0 Δx+2

and then just plug in $0$ to make the value $2$?

Can we prove that $\Delta x$ is continuous?

ZYX
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Prince Z
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  • $\triangle x$ can be understood as simply a notation for a variable. What do you say about the $\lim\limits_{t\to 0}(t+2)$? is the function $f(t)=t$ continuous? – zkutch Dec 09 '23 at 03:56
  • I'm sure f(t)=t is continous, but I think△x is changed value on x axles.. And I also see no one prove that the function is continous before they just plug in. is it just their problem? or I'm someone missing out? – Prince Z Dec 09 '23 at 03:59
  • The proof of continuity $f(t)=t$ comes down to identity between $|t-t_0|$ and $|f(t)-f(t_0)|$, so it is trivial, but maybe some author considers this too. In simple cases often is using fact, that valid arithmetical and composition operations on continuous functions are continuous – zkutch Dec 09 '23 at 04:10
  • thank you very much, but I still really want to know the algebraic proof of that arbitary changes in x axles are continous function because I really want to get intuision of it. – Prince Z Dec 09 '23 at 04:15
  • I found this first answer help by Micheal Hardy https://math.stackexchange.com/questions/1834166/why-doesnt-derivative-difference-quotient-violate-the-epsilon-delta-definition?rq=1 but I have a question that how 0<|h|<δ can be |h|<ε? |h| can be only 0 in later by the density of real number, but It violate the inequalitise in former one. it doest make any sense? – Prince Z Dec 09 '23 at 04:20
  • In continuity definition for $f(t)=t$ for $\forall \varepsilon $ take $\delta = \varepsilon$ and you'll have same inequalities on both sides of implication. – zkutch Dec 09 '23 at 04:21
  • In Michael Spivak - Calculus (2008) on page 116 there is proof of continuity $g(x)=x$. – zkutch Dec 09 '23 at 04:34
  • yeah, I am really sure that f(x) = x, but what I mean is that can we just replace Δx which is (x-c) into just variant h? I think we should prove x-c is continous as well – Prince Z Dec 09 '23 at 05:05
  • Maybe this helps: https://math.stackexchange.com/questions/831126/limit-evaluation-method-inconsistency – MattW Dec 09 '23 at 05:07
  • thanks, you helped a lot – Prince Z Dec 09 '23 at 05:39
  • Replacing $h=x-c$ is actually a composition: if we consider $h=\phi(x)=x-c$, then we have $f(x-c)=f(\phi(x))=f(h)$. Now we came to question when we can replace $\lim\limits_{x\to c}f(\phi(x))$ with $\lim\limits_{h\to 0}f(h)$ and here exists corresponding theorems. – zkutch Dec 09 '23 at 12:48
  • But sometimes it doesn't make sense to plug in $\Delta x = 0$. What do you do with $\dfrac{\sin(\Delta x)}{\Delta x}$? Or many others I could write down. With polynomial functions — which are continuous at $0$ — sure. – Ted Shifrin Dec 10 '23 at 20:06

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