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?
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 $(f'(b_n)) \to -\infty$.