Little Limit

Calculus Level 3

If $f$ is defined on $(0,+\infty )$ and $f',f''$ exists , with $\lim\limits_{x \to +\infty}f(x)=0$ and $f''$ bounded, then what is $\lim\limits_{x\to +\infty}f'(x) ?$

\infty

-\infty

0 1

2 solutions

Logic Alpha
Dec 15, 2020

Let $a>0$ be given.

Let $M_0,~M_1$ and $M_2$ be the supremum of $|f(x)|,~|f'(x)|$ and $|f''(x)|$ on $[a,~\infty)$ respectively.

By Taylor Theorem, there exists $\epsilon\in(x,~x+2h)$ such that $f(x+2h)=f(x)+2hf'(x)+2h^2f''(\epsilon)$ $f'(x)=\frac{1}{2h}[f(x+2h)-f(x)]-hf''(\epsilon)$ $|f'(x)|\leq hM_2+\frac{M_0}{h}$ for any $x>a$ and $h>0$ .

Therefore, $|f'(x)|\leq 2\sqrt{M_0M_2}$ by A.M.-G.M. inequality.

Since $M_2$ is bounded and $\lim_{a \rightarrow \infty}M_0=0$ , $\lim_{a \rightarrow \infty}M_1=0$ , that is, $\lim_{x \rightarrow \infty}|f'(x)|=0$ .

Tom Engelsman
Dec 10, 2020

Use $f(x) = \frac{1}{x}$ for a quick -n- dirty check.

Is $f''$ bounded? $f(x)=e^{-x}$ might be better (its second derivative isn't bounded everywhere, but it is on the interval $f$ is defined over)

Chris Lewis - 6 months ago

Why not taking f(x) =0 constantly? 😉

Daniele Rosmondi - 6 months ago

Even better, Daniele! Me...I just love hyperbolas ;)

tom engelsman - 6 months ago

Little Limit

2 solutions

0 pending reports