Question

 Let 𝑓 be differentiable. Show that if lim 𝑥→∞

𝑓(𝑥) = 𝐿 ∈ ℝ then lim 𝑥→∞

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

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

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

Acording 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

1. It is very easy to give 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.