On Finding The Derivative At Asymptotes

Question

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!

Answer 1

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.