Question

Let 𝑓 be differentiable. Show that if lim𝑥→∞ 𝑓(𝑥) = 𝐿 ∈ ℝ then lim𝑥→∞ 𝑓 ′ (𝑥) = 0. Provided that the latter limit is existing. Give an example where the converse is not true. Also give an example for which the limit of 𝑓 ′ is not existing even though the limit of 𝑓 is the same as given.

Answer 1

Let $f\in L^1(R),\exist\space lim_{x->\infty}f(x)=b\neq0$

Let $\epsilon\space be\space choosen\space as\space0<\epsilon<|b|/2$

According with definition of lim

$\exist M>0:\forall(x>M)|f(x)-b|<\epsilon<|b|/2$

We may think b>0 taking otherwise -f(x)

Then f(x)>b/2 as x>M and $\int\limits_{M}^{M\cdot K} |f(x)| dx>\frac{M\cdot K\cdot b}{2}$

with any K>1.

Also valid that $\int\limits_{-\infty}^{\infty} |f(x)| dx\ge \int\limits_{M}^{M\cdot K} |f(x)| dx>\frac{M\cdot K\cdot b}{2}$

If we will take K rather big we will have

$\int\limits_{-\infty}^{\infty} |f(x)| dx=\infty$  that is controversy

An example $f\in L_1(R)$ such that $\bar{\exist}\space lim_{x->\infty}f(x)$ :

We take  $f(x)=1\space if \space|k-x|\le\frac{1}{k^2},k=1,2,...$ and f(x)=0 otherwise. Then $\int\limits_{-\infty}^{\infty} |f(x)| dx=$$2\cdot \displaystyle\sum_{i=1}^{\infty}\frac{1}{k^2}<\infty$

but $\exists \{x_n\}_{n=1,2,..}$ such that $f(x_n)=1$ if n is odd and $f(x_n)=0$ if n is even there fore limit f(x) on $\infty$ doesn't exist.

if lim𝑥→∞ 𝑓(𝑥) = 𝐿 ∈ ℝ then:

lim𝑥→∞ 𝑓'(𝑥) =lim𝑥→∞ (dL/dx)=0