If Limit of function and derivative exist, then limit of derivative is 0

Question

Any hints for this question , My attempt;
Say $f(x):0$$\rightarrow$$\mathbb{R}$
The by MVT, there exists a $c$$\in$$(0,\infty)$ , such that;
$f'(c)=$$\frac{f(x)-f(0)}{x-0}$
but im not sure about this step..
$\lim_{c \to +\infty}$$f'(c)$=$\lim_{x \to +\infty}$$\frac{f(x)-f(0)}{x-0}$
I am quite sure there is a mistake, any hints would be appreciated.

If $\lim_{x \to +\infty}$$f(x)$ exists and is finite and $\lim_{x \to +\infty}$$f'(x)=b$ then $b=0$.

ok I agree, this might be a duplicate, what are the rules? should I delete this question? — otupygak, Mar 29 '14 at 17:27
No, keep it. People will eventually close it as a duplicate. — Git Gud, Mar 29 '14 at 17:28

score 1 · Answer 1 · answered Mar 29 '14 at 17:23

1

There is this $e^x$ trick. Consider $$f(x)=\frac{e^x f(x)}{e^x}$$ as $x\to\infty$. Use L'Hopital.

answered Mar 29 '14 at 17:23

Pedro

122,002

user2345215 · Answer 2 · 2014-03-29T17:44:24.707

0

$$f(x)=f(a)+\int_a^xf'(t)\ dt$$ So $\displaystyle\int_a^\infty f'(x)\ dx$ converges, because $\lim\limits_{x\to\infty}f(x)$ exists. Buf if $\lim\limits_{x\to\infty}f'(x)$ exists, then it must be $0$, else the integral wouldn't converge.

edited Mar 29 '14 at 17:44

answered Mar 29 '14 at 17:30

user2345215

16,422

the first line is only true when the derivative is a continuous function. – mookid Mar 29 '14 at 17:33
@mookid That's not true. $f$ only has to be differentiable. (this is of course the Newton integral) – user2345215 Mar 29 '14 at 17:34
1

I let you check here: "The Second part is somewhat stronger than the Corollary because it does not assume that f is continuous." – mookid Mar 29 '14 at 17:36
Another interesting point is the absolute continuity definition. – mookid Mar 29 '14 at 17:38
One nice counterexample is the Cantor function, for which $f(1) - f(0) = 1$ but $f'(x) = 0 $ almost everywhere. – mookid Mar 29 '14 at 17:40
@mookid that's not differentiable everywhere. I'm using the Newton integral $\displaystyle(N)!!\int_a^b!f$ which is just by definition $F(b)-F(a)$. – user2345215 Mar 29 '14 at 17:40
@mookid But still has basic properties like if $f\ge g$, then $\displaystyle(N)!!\int_a^b!f\ge (N)!!\int_a^b!g$ if they are both Newton integrable. Therefore my solution is valid. – user2345215 Mar 29 '14 at 17:43
@mookid No comment? – user2345215 Mar 29 '14 at 17:57
I don know anything about eh Newton integral. – mookid Mar 29 '14 at 17:58
@mookid It's extremely simple. If $f=F'$, then $(N)!!\int_a^b!f\overset{\text{def}}{=}F(b)-F(a)$. This is independent of $F$ and it's easy to see that $f\ge0\Longrightarrow(N)!!\int_a^b!f\ge0$. You immediately get that $f>c>0$ where $c\in\mathbb R$ implies $(N)!!\int_a^b!f\ge(b-a)c$ and the conclusion is that if the integral in my answer converges, then $f'\to0$. – user2345215 Mar 29 '14 at 18:20

score 0 · Answer 3 · answered Mar 29 '14 at 17:49

0

$$\limsup_{h\to 0} \left| \frac{f(x+h) - f(x)} h - b \right| = \limsup_{h\to 0} |f'(x+\theta_{x,h} h) - b| \le \limsup_{h\to 0} \sup_{0<\theta<1}|f'(x+\theta h) - b| =0 $$

answered Mar 29 '14 at 17:49

mookid

28,236

If Limit of function and derivative exist, then limit of derivative is 0

3 Answers3