2
$\begingroup$

Let $f:(-\infty,a) \to \mathbb{R}$ be a differentiable function verifying that $\lim\limits_{a^{-}}f = +\infty$. Since the function grows rapidly to infinity when getting close to $a$ in a "finite" space, I wondered if $\lim\limits_{a^{-}} f' = +\infty$. I didn't find any counterexample, so I think the implication is true. Now, for the proof, I thought of fixing an $\bar{a}$ arbitrarily close to $a$, and considering for all $x\in(\bar{a},a)$, the slope : \begin{equation} S:=S_{\bar{a}}(x)=\frac{f(x)-f(\bar{a})}{x-\bar{a}} \end{equation} When $x$ tends to $a$, $S$ goes to infinity and that's for all $\bar{a}$ (so if $\bar{a}$ tends to $a$, it'll be the same), which means that : \begin{equation}\lim\limits_{\bar{a}\to a} \lim\limits_{x \to a} S = +\infty \end{equation} When $x$ approaches $\bar{a}$, S approaches $f'(a)$, and this being true for all $\bar{a}$, we can have : \begin{equation} \lim\limits_{\bar{a}\to a} \lim\limits_{x \to \bar{a}} S = \lim\limits_{\bar{a}\to a} f'(\bar{a})=\lim_{x\to a}f'(x) \end{equation} I feel that the two procedures are equivalent since we always get $x$ and $\bar{a}$ infinitesimally close, so both limits should be equal in some sense. So we get at the end that $f'(x) \to +\infty$.

$\endgroup$

2 Answers 2

6
$\begingroup$

No. Consider $\sin(1/x^2) - 1/x$, $a=0$.

$\endgroup$
6
  • $\begingroup$ There's no need to take the square of $x$. $f(x)=\sin(1/x)-1/x$ will work and makes the proof simpler. $\endgroup$
    – jjagmath
    Jan 2, 2021 at 12:49
  • $\begingroup$ @jjagmath, so you say that showing that $x^{-2}(1-\cos(x^{-1}))$ doesn't converge to $+\infty$ is simpler, then showing, that $x^{-3} (- 2\cos(x^{-2}) - x)$ doesn't converge to $+\infty$ , but you need to show that $\liminf x^{-2}(1-\cos(x^{-1})) < \infty$, so your statement is not true. $\endgroup$ Jan 2, 2021 at 12:58
  • $\begingroup$ $(1-cos(1/x))/x^2$ is $0$ for $x=1/(\pi/2 + n \pi)$ so the limit can't be $\infty$. I would do a similar proof for the function you post, but the calculations are a little more difficult. $\endgroup$
    – jjagmath
    Jan 2, 2021 at 13:09
  • $\begingroup$ @jjagmath, the difficulty in calculating the derivative is the same, but there is an extra step in your solution, because you have to find special type of $x = \frac{1}{\frac{\pi}2 + \pi n}$ and paste them. Maybe you mean that that finding $(x^{-2})'$ is harder, then finding $(x^{-1})'$? Otherwise there's no place, in which your example may be simpler, but in addition it has an extra step. $\endgroup$ Jan 2, 2021 at 13:14
  • 1
    $\begingroup$ May be our perspectives are different. I find my method easier. So I restate my comment: "There's no need to take the square of $x$. $f(x)=\sin(1/x)−1/x$ will work and FOR ME makes the proof simpler." $\endgroup$
    – jjagmath
    Jan 3, 2021 at 0:19
4
$\begingroup$

What you can conclude is that $\lim\sup_{x\to a^{-}}f'(x)=+\infty$. This is because: $\frac{f(x+h)-f(x)}{h}=f'(\xi)$ for some $\xi\in(x,x+h)$ (IVT), so letting $h\to (a-x)^{-}$ makes the left-hand side go to $+\infty$, so there will be points $\xi$ in each interval $(x,a)$ that produce $f'(\xi)$ arbitrarily large.

You cannot, however, conclude $\lim_{x\to a^{-}}f'(x)=+\infty$. One can found a counterexample by tweaking a well-known function, e.g. $f(x)=-\frac{1}{x}$ with $a=0$.

Take a sequence $x_n\to a^{-}$ monotonously, e.g. $x_n=-\frac{1}{n}$. Now make the sequence $y_n$ as follows:

  • For odd $n$, set $y_{2n-1}=f(x_{2n-1})$, i.e. in our example: $y_{2n-1}=2n-1$.
  • For even $n$, keep the same value, i.e. $y_{2n}=y_{2n-1}=f(a_{2n-1})$, i.e. in our example: $y_{2n}=2n-1$.

Now all it takes is to pass a smooth curve through the points $(x_n, y_n)$. This can be done in many ways - let your imagination go wild! You get a differentiable function $g(x)$ such that $g(x_n)=y_n$.

The function $g$ looks like smoothed staircase which climbs towards $+\infty$ at about the same rate as $f$: in fact, they coincide at points $x_{2n+1}$. However, $g$ has also got those intervals $(x_{2n-1},x_{2n})$ where it is (approximately) "flat".

This lets us conclude that, on each $(x_{2n-1}, x_{2n})$, there is a point $\xi\in(x_{2n-1}, x_{2n})$ such that $f'(\xi)=0$ (Rolle theorem). As $x_{2n-1}\to a^{-}$ as $n\to\infty$, we conclude that $\lim\inf_{x\to a^{-}}f'(x)\le 0$. However, because $\lim\sup_{x\to a^{-}}f'(x)=+\infty$ (proven above), we conclude that $\lim_{x\to a^{-}}f'(x)$ does not exist.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .