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A proof offered for the derivative of a constant is shown here:

https://tutorial.math.lamar.edu/classes/calci/DerivativeProofs.aspx

enter image description here

My question is: why is it "permitted" to indicate:

$$\lim_{h\to0} [f(x+h)] = c$$

It seems like a step is being skipped by implicating that $f(x)+h$ is actually $f(x)$ [ presumably since the function itself is a constant?]

WestCoastProjects
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  • $f(x+h)$=c since $f(x)=c$ is constant – Sine of the Time Feb 11 '23 at 21:00
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    There is no $f(x)+h$ in your screenshot. There is even no $\lim_{h\to0} f(x+h)= c.$ There is only an expression containing $f(x+h),$ equal to the same expression where $f(x+h)$ has been replaced by $c.$ – Anne Bauval Feb 11 '23 at 21:02
  • @AnneBauval yea - i just corrected that – WestCoastProjects Feb 11 '23 at 21:04
  • @SineoftheTime Your comment nails my question: i had not been interpreting correctly. feel free to make an answer if not "beneath you" in terms of being so simple – WestCoastProjects Feb 11 '23 at 21:05
  • (Follow up) There is only an expression containing $f(x+h),$ equal to the same expression where $f(x+h)$ has been replaced by $c.$ This is legitimate because $f$ of anything is equal to $c.$ Same for the replacement by $c$ of the other subexpression $f(x).$ – Anne Bauval Feb 11 '23 at 21:08
  • @WestCoastProjects you can visualize graphically for example: $f(x+h)-f(x)$ is an increment long the $y$-axis, but since $f$ is constant, the increment is $0$ as $f(x)=c$ is parallel to the $x$-axis – Sine of the Time Feb 11 '23 at 21:09
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    @AnneBauval That's a reasonable explanation as well - either or both of you or SineoftheTime are invited to make your comment an answer. – WestCoastProjects Feb 11 '23 at 21:13

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There is no "$\lim_{h→0}f(x+h)=c$" in your screenshot. There is only an expression containing $f(x+h),$ equal to the same expression where $f(x+h)$ has been replaced by $c.$ This is legitimate because $f$ of anything is equal to $c.$ Same for the replacement by $c$ of the other subexpression $f(x).$

Anne Bauval
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