# Boundary condition on a function $f(x)$ and its first derivative at $x\rightarrow\pm\infty$

If a continuous function $f(x)$ of a real variable $x$ is such that $f(x)\rightarrow 0$ as $x\rightarrow \pm \infty$, does it necessarily mean that $\frac{df}{dx}\rightarrow 0$ as $x\rightarrow \pm \infty$? If yes, can I prove this?

If not, what is a counter-example where this is not true. The examples, I can think of $\frac{1}{x}, e^{-x}$ etc satisfy this.

• By the way, if you want this to be true, you need the requirement of monotonicity as $x\to\pm\infty$. – Simply Beautiful Art Jan 12 '17 at 19:07
• It's still not true with monotonicity @SimpleArt. See my post. There's also an explicit construction here: math.stackexchange.com/questions/788813/… – Kaj Hansen Jan 12 '17 at 20:10
• For this to be true you need $f''$ bounded. – MathematicsStudent1122 Jan 12 '17 at 21:41

Interestingly, this is not even true if we add a monotonicity hypothesis on the function. I don't know an explicit formula for this off the top of my head, but consider something like the following:

Essentially the function is constant between integer-values of its domain, and at each integer value, it takes a steep decline. The distance the function "drops" at each integer value limits to zero in such a way that we have $\displaystyle \lim_{x \rightarrow \infty} f(x) = 0$. However, because there always will be a steep drop at each integer value (albeit arbitrarily small drops), the derivative cannot limit to zero. Note that the "drops" can be smoothed out, say with something bump-function-esque, to make $f$ infinitely differentiable.

No. Consider the following:

$$f(x)=\frac{\sin(e^x)}x$$

as $x\to\pm\infty$, $f(x)\to0$. However,

$$f'(x)=\frac{xe^x\cos(e^x)-\sin(e^x)}{x^2}\to\text{DNE as }x\to+\infty$$

Here is the graph:

• very nice example! Do you mind adding a plot of the function to illustrate that the function goes more and more crazy in a smaller and smaller interval (on the y-axis) as $x\to\infty$? – Surb Jan 12 '17 at 19:00
• @Surb Done I do think :-) – Simply Beautiful Art Jan 12 '17 at 19:01
• :) thanks, also if you want the divergence on both sides you might consider $\sin(e^{x^2})/x$ – Surb Jan 12 '17 at 19:03
• @Surb True true... taking advantage that all you really need to do is make my function symmetric. – Simply Beautiful Art Jan 12 '17 at 19:04
• Actually, the limit isn't $\infty$, it doesn't exist. – Michael Hoppe Jan 12 '17 at 19:04