13
$\begingroup$

Given that $f$ is differentiable, what does $\lim\limits_{x \to \infty} f(x) = 1$ say about $\lim\limits_{x \to \infty} f^\prime(x)$ ? Intuitively I feel that it's $0$.

I attempted to solve this by trying to evaluate $\lim_{x \to \infty} \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $ by trying to interchange the limits after showing that $\frac{f(x + h) - f(x)}{h}$ converges uniformly as $x \to \infty$. But I couldn't proceed further.

Trying to work backwards, as a specific example, $f(x) = \arctan(x)$ came to my mind. It's derivative certainly goes to $0$ as $x \to \infty$. Doesn't this show that $\frac{f(x + h) - f(x)}{h}$ converges uniformly to $0$ as $x \to \infty$ ?

I'm totally confused! I would really appreciate if anyone told me what I am doing wrong.

  • 0
    @El I couldn't see how I could show uniform convergence. So I tried to work backwards using a specific example to see if uniform convergence would come up.2011-04-29

2 Answers 2

7

This is in response to Balaji's comment.

Lemma. Let $f\colon (a,+\infty)\to\mathbb{R}$ be a bounded $C^1$ function such that \lim_{x\to+\infty}f'(x)=\delta exists. Then $\delta=0$.

Proof. We proceed by contradiction and assume that $\delta\ne0$. Suppose first that $\delta>0$. Then there exists $x_0>a$ such that f'(x)\ge\delta/2 for all $x\ge x_0$. Then f(x)=f(x_0)+\int_{x_0}^xf'(t)dt\ge f(x_0)+\frac{\delta}{2}(x-x_0)\quad\forall x\ge x_0. In particular $f$ is unbounded. The case $\delta<0$ is done similarly.

If you do not want to use integrals, you may reason as follows. Let $g(x)=f(x)-f(x_0)-\frac{\delta}{2}(x-x_0).$ Then $g(x_0)=0$ and g'(x)\ge 0 for all $x>x_0$. It follows that $g$ is increasing in $(x_0,+\infty)$ and $g(x)\ge 0$ for all $x>x_0$.

24

Nothing. you can take something like: $x\mapsto \frac{\sin(x^2)}{x}+1$.

  • 0
    @Julián Thanks! After seeing the answers above I had reached this conclusion, but was waiting for someone to say it. But how can one show it ? If you can make an answer out of it, I'll be very glad.2011-04-30