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How do you prove that the derivative of a continuous function which approaches infinity from the right or the left of some point $x_0$ also approaches infinity? I suppose the definition of the limit needs to be applied somehow, but I don't know how. Thank you.

$\lim_{x\rightarrow x_0^+}{f(x)} = \infty \rightarrow \lim_{x\rightarrow x_0^+}{f'(x)} = -\infty$

Is $$f'(x)=-\dfrac{1}{x-x_0}(\sin(\dfrac{1}{x-x_0})+1)$$ a counterexample or not?

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  • $\begingroup$ This is false. You cannot prove it. $\endgroup$ – Crostul Jan 6 at 19:23
  • $\begingroup$ Could you clarify why? $\endgroup$ – Andrew Blitz Jan 6 at 20:03
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    $\begingroup$ @AndrewBlitz: Assuming there is a derivative it is true that $\liminf_{x \to x_0+} f'(x) = -\infty$ but I believe it is possible that the limit may not exist. $\endgroup$ – RRL Jan 6 at 20:12
  • $\begingroup$ What if $f'(x) = \dfrac{1}{x-x_0}(sin(\dfrac{1}{x-x_0})+1)$? $\endgroup$ – Andrew Blitz Jan 6 at 22:22
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I doubt that the statement is true. For example, $$ f(x) = \int_x^1 \frac{1+\cos(\frac{\pi}{t})}{t}dt $$ has vanishing derivative at $x=\frac{1}{(2k+1)}$, $k\ge 1$ but for all $\epsilon <\frac{1}{N}$, we have $$ f(\epsilon)\ge\int_{\frac{1}{N}}^1 \frac{1+\cos(\frac{\pi}{t})}{t}dt = \int_{1}^N \frac{1+\cos(\pi u)}{u}du\ge \sum_{n=1}^{N-1} \frac{1}{n+1}\int_{n}^{n+1}[1+\cos(\pi u)] du =\sum_{n=1}^{N-1}\frac{1}{n+1}, $$which is saying that $\lim_{x\to 0^+} f(x) =\infty.$

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It is true that $\liminf_{x \to x_0+} f'(x) = - \infty$ but it is not necessary that $\lim_{x \to x_0} f'(x)$ exists.

To prove the first statement take a sequence $x_n \to x_0+$ where $f(x_n) \to +\infty$. Take a fixed $y$ such that $x_0 < x_n < y$ and $f(x_n) > f(y)$ for all $n$. By the MVT there is a sequence $\xi_n$ such that

$$f'(\xi_n) = \frac{f(y) - f(x_n)}{y - x_n} < \frac{f(y) - f(x_n)}{y - x_0}$$

and the RHS converges to $-\infty$ as $n \to \infty$.

For a counterexample to existence of the limit, take $x_0 = 0$ and $f(x) = - \log(x^2 \sin \frac{1}{x})$. Here we have $f(x) \to +\infty $ as $x \to 0+$, and

$$f'(x) = \frac{2}{x} - \frac{\cot \frac{1}{x}}{x^2} = \frac{1}{x^2}\left(2x - \cot \frac{1}{x} \right)$$

Although $f'$ is unbounded in a neighborhood of $0$, it oscillates perpetually (passing through $0$) and the limit does not exist. Note that $\cot y - 2/y$ has infinitely many zeros in $(1, \infty)$.

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  • $\begingroup$ Good catch! Sorry I missed it. $\endgroup$ – Ben W Jan 7 at 1:01
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You need f to be differentiable and for the limit to exist. For $x_0<x<x+1$ select $x_0<y<x$ with $f(y)>f(x)+n$. Apply the mvt and let $n\to\infty$. Then we get $\liminf_{x\to x_0^+}f'(x)=-\infty$, which since the limit exists means $\lim_{x\to x_0^+}f'(x)=-\infty$.

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    $\begingroup$ Could produce a few more steps to show that this proves $\lim_{x \to x_0+}f'(x) = -\infty$ and not just the existence of a sequence $\xi_n \to x_0+$ where $f'(\xi_n) \to -\infty$ (i.e., unbounded but with non-existent limit) $\endgroup$ – RRL Jan 6 at 20:10
  • $\begingroup$ @RRL oops! Fixed. $\endgroup$ – Ben W Jan 7 at 1:02

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