if function "derivative" is infinite somewhere, is it necessarily continuous around that point?

Question

if $ \lim_{x \to a^{+}} \frac{f(x)-f(a)}{x-a} = +\infty $ , is the function f continuous on $ (a, a+\epsilon) $ for some small $\epsilon>0 $ ? if that's the case, can someone please provide with a proof relying on $\epsilon \delta$ definition of limits and continuity and really basic knowledge of real analysis (i just started real analysis on my own, not a math major so i don't know many theorems), Thank you in advance

Answer 1

Assume $a=0$. Take any function $h\geq 0$ such that $h(0)=0$. Then $f(x)=h(x)+ \sqrt{x}$ will satisfy the hypothesis $$ \lim_{x\to 0^+}\frac {f(x)-f(0)}{x}\geq\lim_{x\to 0^+}\frac {\sqrt x}{x} =+\infty. $$ If in addition $h$ is not continuous on any interval $(0,\epsilon)$, then $f$ isn't as well. So the question becomes: can you think of a function $h\geq 0$ that is not continuous on any interval $(0,\epsilon)$, $\epsilon>0$? It should be easier for you to find one. Tell me if you need a hint on that.

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Edit. The above limit is infinite because, since $f(x)\geq\sqrt x$, you have $$ \liminf_{x\to 0^+}\frac{f(x)}{x}\geq\liminf_{x\to 0^+}\frac{\sqrt x}{x}=\liminf_{x\to 0^+}x^{-1/2}=+\infty. $$ Since the limsup is always greater than the liminf, the limsup is $+\infty$ as well, so that $$ \liminf_{x\to 0^+}\frac{f(x)}{x}=\limsup_{x\to 0^+}\frac{f(x)}{x}=+\infty, $$ so the limit exists and is $+\infty$.

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Comment. Morally, prescribing the derivative at some point $a$ tells you nothing on the continuity of the function around $a$.

A related example is the following. Choose any function $0\leq f(x)\leq x^2$ on the interval $(-1,1)$. Then, you can check that $$ 0=\liminf 0\leq\liminf_{x\to 0}\frac{f(x)-f(0)}{x-0}\leq\limsup_{x\to 0}\frac{f(x)-f(0)}{x-0}\leq\limsup_{x\to 0}\frac{x^2}{x}=\lim_{x\to 0}x=0, $$ so that all the above quantities are zero, and $f'(0)=0$. So $f$ is differentiable (hence continuous) at $x=0$, but it is surely possible that such an $f$ is discontinuous at any other point. For instance, take $h$ to be the function equal to $0$ over the rationals and $1$ over the irrationals, and set $f(x)=x^2\cdot h(x)$. Then $h(x)=f(x)\cdot\frac{1}{x^2}$ and $\frac{1}{x^2}$ is continuous for any $x\neq 0$, so $f$ must be discontinuous at any given point $x\neq 0$ (otherwise $h$ would be continuous as a product of continuous functions).

Answer 2

Define $f$ on ${\Bbb R}^+$ as follows, $$f(x)=\cases{\frac1x&if $\frac1x$ is not an integer\cr \frac1x+1&if $\frac1x$ is an integer,\cr}$$ together with $f(0)=0$. Then for every positive integer $n$ we have $$\lim_{x\to1/n}f(x)=n\ ,\quad f\Bigl(\frac1n\Bigr)=n+1\ ,$$ so $f$ is discontinuous at $x=\frac1n$. But for the difference quotient at $0$ we have $$\frac{f(x)-f(0)}{x-0}=\frac{f(x)}{x}\ge\frac1{x^2}\to\infty\quad\hbox{as}\quad x\to0^+\ .$$