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I have 2 questions here. First. Why can't we find the derivative of a function at the vertical asymptote?

Second. Can we find the derivative a function at it's horizontal asymptote, and if so/not, why? And if so, is the derivative 0?

On the first question:

  1. I know you can't differentiate if the function is not continuous. But is not the function defined at the asymptote, and therefore continuous (because the values on both sides also approach that definition)? At the asymptote, the function is infinity! It's not a real number, but it's still a definition! So why must the definition of it be a real number? Can't we just use infinity, and say that the derivative of the function at the vertical asymptote is infinity?

On the second question:

  1. Can one differentiate at the horizontal asymptote of a function? I know the horizontal asymptote isn't reached by any real number, but it is at x equals infinity. So does this count as the function being defined at the horizontal asymptote? And if so/if not, why?

  2. And if it is defined there, is the derivative 0? If so/if not, why?

Is it possible to give your answer without epsilon delta proofs, at the level of someone learning Khan Academy Calculus? Thank you so much!

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    $\begingroup$ Consider the function $y=1/x$, which has a vertical asymptote at $x=0$. As $x$ approaches $0$, the value of $y$ approaches either positive and negative infinity, so saying "the function is infinity" is somewhat inappropriate. In any case, if you want to include infinite (and infinitesimal) quantities among the inputs and outputs of functions, you'll want to look into "nonstandard calculus". $\endgroup$ – Blue Sep 2 '18 at 14:56
  • $\begingroup$ $y=1/x^2$ avoids the complication I mention, but saying "the function is infinity" at $x=0$ remains problematic. "Standard Calculus" is about the real numbers, and only the real numbers; the game is played with that restriction. There are advantages to including infinities in the mix (see the extended real numbers), but the ramifications are significant. (For instance, the basic rules of arithmetic break.) "Non-standard Calculus" addresses the nuances of this, and is a legitimate topic of study, but it's a separate one. $\endgroup$ – Blue Sep 2 '18 at 15:26
  • $\begingroup$ Well, is the function defined at the $x$-coordinate of its asymptote? Usually not. $\endgroup$ – amd Sep 2 '18 at 20:39
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Let me try to clear your misconception. Let us consider a function $f(x)=\frac{1}{x}$. The graph would be something like this. There's an assymptote at $x=0$, it means that the function is not defined at $x=0$. But $\lim_{x\to0}f(x)=\infty$, so you are confusing between $\lim_{x\to0}f(x)$ and $f(0)$, which are two different things (as, in this case $f(0)$ doesn't exist).

ANS 1

  • derivative doesn't exist as f(x) is not continuous at x=0.

ANS 2

  • $\lim_{x\to \infty}f(x)=0$, but $\infty$ itself is not a concrete definition it's up to your imagination to find it's value so, $f(\infty)$ itself doesn't make sense (but $\lim_{x\to \infty}f(x)$ does), it means as $x\rightarrow \infty$, $f(0)\to 0$ but never becomes equal to $0$. Hope this helps.
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    $\begingroup$ Welcome to the site! Here one uses markdown as syntax and LaTeX for math only. This improves readability for both humans and computers (parsing by computers, that is). I took the liberty to edit the formatting of the answer (but, hopefully, did not change the content). $\endgroup$ – Dirk Sep 4 '18 at 11:26
  • $\begingroup$ @Dirk I was unable to insert the link. Thanks for help. Now I can do that.;-) $\endgroup$ – Subhajit Halder Sep 4 '18 at 11:28

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