# How do you prove the following about a function approaching infinity at some point?

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?

• This is false. You cannot prove it. – Crostul Jan 6 at 19:23
• Could you clarify why? – Andrew Blitz Jan 6 at 20:03
• @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. – RRL Jan 6 at 20:12
• What if $f'(x) = \dfrac{1}{x-x_0}(sin(\dfrac{1}{x-x_0})+1)$? – Andrew Blitz Jan 6 at 22:22

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 $$\lim_{x\to 0^+} f(x) =\infty.$$

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 $$(1, \infty)$$.

• Good catch! Sorry I missed it. – Ben W Jan 7 at 1:01

You need f to be differentiable and for the limit to exist. For $$x_0 select $$x_0 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 $$\lim_{x\to x_0^+}f'(x)=-\infty$$.

• 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) – RRL Jan 6 at 20:10
• @RRL oops! Fixed. – Ben W Jan 7 at 1:02