Let $f:[0,\infty[ \rightarrow \mathbb{R}$ be differentiable on $(0,\infty)$ such that $f'(x) \to b \; as \; x \to \infty$

Question

(a) Show that for any $h>0$ we have $\lim_{x \to \infty} \frac{f(x+h) - f(x)}{h}=b$

(b) Show that if $f(x) \to a $ as $x \to \infty$ then $b = 0$

(c) Show that $\lim_{x\to \infty} \frac{f(x)}{x}=b$

I have no clue on how to approach this problem... I tried to show $h \to 0$ but I could not.

Hint for (a) https://en.wikipedia.org/wiki/Mean_value_theorem. — eranreches, Dec 09 '17 at 00:40
@P.Koymans: why do you think it is wrong. It should be obvious that it is true. — Paramanand Singh, Dec 09 '17 at 02:24
@P.Koymans $\lim_{x\to\infty}f(x) \to a\implies \lim_{x\to\infty}f(x)=\lim_{x\to\infty}f(x+k),\forall k\in\Bbb R\implies \lim_{x\to\infty} f'(x)=0$. in $f(x)=x$ we have $\lim_{x\to\infty}f(x)\ne a$(you have to remember that $\infty$ is not a number!!) — ℋolo, Dec 09 '17 at 02:32
@Holo It was wrong, but changed later. See the edit history. — P. Koymans, Dec 09 '17 at 12:00

user284331 · Accepted Answer · 2017-12-09T00:50:45.487

1

For $\epsilon>0$, choose $M>0$ such that $|f'(u)-b|<\epsilon$ for $u\geq M$. Now fix an $h>0$, then for all $x>M$, we have $\left|\dfrac{f(x+h)-f(x)}{h}-b\right|=|f'(\xi_{x,h})-b|<\epsilon$, this proves a).

Now assume the result in a), put $h=1$, for some $N>0$, we have $|f(x+1)-f(x)-b|<\epsilon$ for $x\geq N$. Now taking $x\rightarrow\infty$, we have $|b|=|a-a-b|\leq\epsilon$, this is true for all $\epsilon>0$, so $b=0$.

edited Dec 09 '17 at 00:50

answered Dec 09 '17 at 00:43

user284331

55,591

For c), I have answered someone else in the following link. – user284331 Dec 09 '17 at 00:45
https://math.stackexchange.com/questions/2548583/how-to-prove-that-lim-limits-x-to-infty-fx-l – user284331 Dec 09 '17 at 00:45
for (a) don't I have to first show that $h \to 0$ because only then $f'(x) = \lim_{h \to 0 } \frac{f(x+h)- f(x)}{h}$ – So Lo Dec 09 '17 at 00:57
I don't follow? In a), the limit taken is about $x\rightarrow\infty$, not $h\rightarrow 0$. – user284331 Dec 09 '17 at 00:57
You are using $f'(x) = \frac{f(x+h)-f(x)}{h}$ in your first inequality am I right ? but isn't that true only if $h \to 0$ – So Lo Dec 09 '17 at 01:02
I don't use $f'(x)=\dfrac{f(x+h)-f(x)}{h}$ and surely this inequality is not true. – user284331 Dec 09 '17 at 01:03

Let $f:[0,\infty[ \rightarrow \mathbb{R}$ be differentiable on $(0,\infty)$ such that $f'(x) \to b \; as \; x \to \infty$

1 Answers1