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According to wikipedia (https://en.wikipedia.org/wiki/Vertical_tangent): A function ƒ has a vertical tangent at x = a if the difference quotient used to define the derivative has infinite limit:

${\displaystyle \lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}={+\infty }\quad {\text{or}}\quad \lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}={-\infty }.}$

And it is possible to often find detect a vertical tangent by taking the limit of the derivative such that:

${\displaystyle \lim _{x\to a}f'(x)={\pm\infty }{\text{,}}}$

My question is when is this not possible. This these follow the limit definition of a derivative, what is a possible function that has a vertical asymptote that can't be computed by the derivative or such that the derivative does not provide any information about the vertical tangent.

My logic is that if the limit:

${\displaystyle \lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}={+\infty }\quad {\text{or}}\quad \lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}={-\infty }.}$

simply exists then the derivative is computable. Can someone correct me on my understanding?

UT-HJ
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  • $\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ and $\lim_{x\to a}f'(x)$ are exactly the same thing. – K.defaoite Dec 21 '22 at 03:53
  • Added the link to the Wikipedia specifying that this is only often the case for vertical tangents – UT-HJ Dec 21 '22 at 03:58
  • I think they have misused the word "often" – K.defaoite Dec 21 '22 at 04:01
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    Perhaps it is do with the fact that a vertical tangent exists only if the function itself exists; for example there exists a vertical tangent to $y=\sqrt{x}$ at $x=0$, however, $y=\frac{1}{x}$ does not have a vertical tangent at $x=0$ (as the curve is undefined here), even though setting the derivative to "undefined" gives $x=0$. – user1050148 Dec 21 '22 at 04:23
  • @K.defaoite what if the function is not continuous? is the difference quotient still computable – UT-HJ Dec 21 '22 at 04:38
  • @UT-HJ I cannot think of a simpler example, for now. But you can have a look at https://math.stackexchange.com/questions/3332275/continuity-of-a-strange-function/3332351#3332351 – dfnu Dec 21 '22 at 06:33
  • @K.defaoite your assertion is false in general. You may well have a derivative without the limit of $f'(x)$ existing. Take for instance the usual $f(x) = x^2\sin(1/x)$ with removed discontinuity in $0$. – dfnu Dec 21 '22 at 06:35
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    @K.defaoite You are claiming that $f'(a)$ and $\lim_{x\to a}f'(x)$ are exactly the same thing, so you are claiming that derivatives are always continuous. This is far from the case, however. – Trebor Dec 21 '22 at 16:18
  • Okay, my mistake. – K.defaoite Dec 21 '22 at 23:03

1 Answers1

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In general (recall De l'Hospital's Theorem) it is possible that $\lim_{x\to a}f'(x)$ does not exist and yet $\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ exists (finite or infinite). Of course if the former exists it has to be equal to the latter.

Consider,e.g., $$f(x) = \begin{cases} \sqrt[3]{x} +x\cos\left(\frac1x\right) & (x\neq 0)\\ 0 & (x=0) .\end{cases}$$

Here $$\lim_{x\to 0} \frac{f(x)}{x} = \lim_{x\to 0} \left[\frac1{\sqrt[3]{x^2}} + \cos\left(\frac1x\right)\right]=+\infty.$$

However, for $x\neq 0$, $$f'(x) = \cos\left(\frac1x\right) + \frac1x\sin\left(\frac1x\right) + \frac1{3\sqrt[3]{x^2}},$$ and $\lim_{x\to 0} f'(x)$ does not exist, as you can check by testing the two null sequences $a_n = \frac1{\frac{\pi}2+2\pi n}$ and $b_n = \frac1{\frac{3\pi}2+2\pi n}$, leading to $(f'(a_n)) \to +\infty$ and $(f'(b_n)) \to -\infty$.

enter image description here

dfnu
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  • Neat example! However, to avoid giving the OP the wrong impression, it's worth pointing out that a sketch such as the graph you've given clearly shows that such examples exist without the need for algebraic/analytic calculations, unless an example defined in some specific algebraic/analytic manner is desired or the existence of an example defined explicitly in terms of elementary functions is desired. – Dave L. Renfro Dec 21 '22 at 16:02
  • Hello @DaveL.Renfro ! If I interpret correctly what you are saying: De l'Hospital Theorem shows that the existence of the limit $f'(x)$ is a sufficient but not necessary condition for the existence of the limit of the difference quotient, as I pointed out at the beginning of the answer, so that it is incorrect to state that the two limits are in fact the same. Or maybe I misunderstand your remark? – dfnu Dec 21 '22 at 16:10
  • I'm just saying that the graph itself makes it clear that such an example exists, so there is no need for analytical limit calculations, but you should keep them anyway just to show how one can carry them out. I'm only mentioning this because beginning students often think they have to come up with a specific formula (usually only using "precalculus" functions) to determine something like this. For example, instead of the oscillating curve you've drawn between the upper and lower boundaries, you could also have a staircase type function that stays within these boundaries, (continued) – Dave L. Renfro Dec 21 '22 at 20:24
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    which shows that as you approach $0$ you can also have the derivative equal to zero infinitely often (even be zero throughout entire intervals) and also the derivative always being nonnegative (note in your example the derivative is negative, even arbitrarily large negative, infinitely often as you approach $0).$ Of course, this property that the derivative is infinitely often zero and always nonnegative might not be of any interest, but note how easy it is to draw such an example, whereas finding a formula for it will likely be difficult and produce something too messy to easily interpret. – Dave L. Renfro Dec 21 '22 at 20:24