Why do we say limits which go to infinity are not defined?

Question

I recently saw this article by oregonstate.edu about the differentiability of the function $y=\sqrt{X}$ at $x=0$ they claim that the derivative at $x=0$ does not exist.i understand the concept of one-sided differentiability and how the $\lim_{h \to 0-} \frac{\sqrt{x+h}-\sqrt{x}}{h} $ does not exist but for values of greater than 0 and approaching from the right they say that the limit does not exist as it is positive infinity. my question is why isn't this limit defined surely we know it is approaching infinity so why can't we say it has an infinite slope

Answer 1

This question gets into subtleties of how we define derivatives in terms of unidirecional limits. Let's consider the outcome in two different number systems.

Real analysis

The derivative exists iff two one-sided derivatives exist with equal value. The right-derivative$$\left.\lim_{h\to0^+}\frac{\sqrt{x+h}-\sqrt{x}}{h}\right|_{x=0}=\lim_{h\to0^+}\frac{\sqrt{h}}{h}=\color{blue}{+\infty}$$ is an extended real number; the left-derivative $\lim_{h\to0^-}\frac{\sqrt{h}}{h}$ does not exist.

Complex analysis

$f^\prime(0)$ exists with value $L$ iff $\lim_{r\to0^+}\frac{f(re^{i\theta})-f(0)}{re^{i\theta}}=L$ for all $\theta\in\Bbb R$. For $f(z)=z^{1/2}=e^{\tfrac12\ln z}$,$$\lim_{r\to0^+}\frac{f(re^{i\theta})-f(0)}{re^{i\theta}}=\lim_{r\to0^+}\frac{\sqrt{r}e^{i\theta/2}}{re^{i\theta}}=\color{red}{\infty}$$is an extended complex number. Do not confuse $\color{blue}{+\infty}$ with $\color{red}{\infty}$.

One more point: if we'd been differentiating $x^{1/3}$ at $x=0$, the two-sided derivative would again not exist in real analysis, but this time for a different reason: the left-derivative would exist, but would be $-\infty$ rather than $+\infty$.