# Is $\lim\limits_{a^{-}}f = +\infty \Rightarrow \lim\limits_{a^{-}}f' = +\infty$ true? If so, proof?

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 $$f'(x) \to +\infty$$.

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

• There's no need to take the square of $x$. $f(x)=\sin(1/x)-1/x$ will work and makes the proof simpler. Jan 2 at 12:49
• @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. Jan 2 at 12:58
• $(1-cos(1/x))/x^2$ 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. Jan 2 at 13:09
• @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. Jan 2 at 13:14
• 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." Jan 3 at 0:19

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.