Derivative Counterexamples - Calculus

Question

I need counterexamples for the following (I guess these claims are not correct):

If $ lim_{n\to \infty} n\cdot (f(\frac{1}{n}) - f(0) ) =0$ then $f$ is differentiable at $x=0$ and $f'(0)=0$ .
If f is defined in a neighberhood of $a$ including $a$ and differentiable at a neighberhood of $a$ (except maybe at $a$ itself), and $lim_{x\to a^- } f'(x) = lim_{x\to a^+} f'(x) $ , then $f$ is differentiable at $x=a$.
If $f$ is diff for all $x$ and satisfies $lim_{x\to \infty } f'(x) =0 $ then there exists a number $L<\infty$ for which $ lim_{x\to \infty} f(x)= L$
If $f$ is diff for all $x$ and satisfies $lim _{x\to \infty} f(x)= L $ then $lim_{x\to\infty} f'(x)=0 $ .
If $f $ is diff at $x=0$ and $lim_{x\to 0 } \frac{f(x)}{x} =3 $ , then $f(0)=0$ and $f'(0)=3$

Thoughts:

5) I think this claim is correct and follows from the uniqueness of the derivative... I have no idea how to prove it, but it sounds reasonable

3) Isn't a counterexample for this is $f(x)=lnx$ ?

4) I have tried using some trigonometric functions, but still couldn't manage to find a counterexample

2) I guess that an example for this would be a function that its derivative isn't defined at this point , but its limits do

1) have no idea... It sounds incorrect (although I guess that the other direction of the claim is correct)

Help?

Thanks !

For 5., use the definition of derivative: $f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}x$, so if $f(0)=0$, we are done. But $f(0)=\lim_{x\to0}f(x)=\lim_{x\to0}x\cdot\frac{f(x)}x=0\cdot3=0$. Here, I used that $f(0)=\lim_{x\to0}f(x)$, which means that $f$ is continuous at $0$, which we know is true since $f$ is differentiable at $0$, and differentiability entails continuity. — Andrés E. Caicedo, Dec 13 '13 at 08:02
For 4 consider something like $f(x)=\frac{\sin(e^x)}{x^2+1}$. — Eric, Dec 13 '13 at 08:39
@Eric : Can you please explain why does that limit of your example doesn't go do zero ? The derivative of what you mentioned goes to infinity ? Thanks! — criticism, Dec 13 '13 at 17:35
For the function, $f(x)$, that I gave above, $\lim_{x\to\infty}f(x)=0$. If you don't want it to be zero just add $L$ to it. The derivative is $\frac{e^x\cos(e^x)(x^2+1)-\sin(e^x)2x}{(x^2+1)^2}$ so the limit of $f'(x)$ does not exist. It might be easier to consider something like $\frac{\sin(x^3)}{x^2+1}$. We use this denominator because it is non-zero for all $x\in\mathbb{R}$. — Eric, Dec 13 '13 at 20:52

score 2 · Answer 1 · answered Dec 13 '13 at 08:24

2

There is a lot of space between $1/n$
The only place to go wrong is at $x=a$. Break it there.

answered Dec 13 '13 at 08:24

Empy2

50,853

I know. this is why I guess the claim is incorrect. But I can't find any proper counterexample that will allow the limit to be zero if I plug in $\frac{1}{n}$ , but won't be zero for other values...
I tried using $f(x)=x^2, x\neq 0 $ and $f(0)=4$ , but it doesn't work, because the derivative will be 0 in any case... will you please help me with these 2? THanks !

criticism

Dec 13 '13 at 17:51

mathematics2x2life · Answer 2 · 2013-12-13T07:57:51.393

1

$5$. Think, if you take the limit you get $\frac{0}{0}$, what can you use then?

$4$. What would it mean if the derivative were positive or negative 'all the way' to infinity?

$3$. Your example works just fine since $\log x$ is differentiable for all $x$ in its domain, it's slopes go to $0$, and the function grows without bound.

$2$. If a function is differentiable, then it is continuous. Think of the parabola $f(x)=x^2$ at the origin, think about the slopes as you approach $0$ from the left/right. Is there something you could do to the parabola at $0$ to make the statement untrue?

$1$. Think derivative definition.

edited Dec 13 '13 at 07:57

answered Dec 13 '13 at 07:50

mathematics2x2life

13,322

If I get you right, according to your first argument $\lim_{x\to 0}\frac{3x}{3}=\frac{0}{0}\neq 3$. – flonk Dec 13 '13 at 08:29
I don't quite understand what you are trying to motivate with your hint for number four. Can you expand on that a bit? – Eric Dec 13 '13 at 08:51
@flonk You don't follow. Think l'Hôpital's rule. – mathematics2x2life Dec 13 '13 at 16:18
@Eric If the derivative is nonzero, the function is either...or.... . So the function does what? Moreover, if the derivative 'cycles' causing the function to oscillate, can the function even approach a limit, this means.... – mathematics2x2life Dec 13 '13 at 16:19
A function with an oscillating derivative can approach a limit. Consider the example I gave in a comment to the original question. – Eric Dec 13 '13 at 17:04
@Eric I wasn't trying to say that. The case where the derivative oscillates clearly prevents $\lim f'(x)=0$, unless the derivative both oscillates and decreases. That would need to be a separate consideration. – mathematics2x2life Dec 13 '13 at 17:21
@mathematics2x2life : I think I managed with everything, except for 1 and 2 .
1. I guess it is incorrect ... but can't find any counterexample.
2. I want to put the value of $f'(0)$ different from the one of $f'(x)$ obviously... But defining the function differently at $0$ doesn't work (see my comment above)
Will you help ? Thanks !
– criticism Dec 13 '13 at 17:53
@criticism For $2$, that is not the idea. Take the ordinary parabola except at $x=0$ have the value of the function by $1$. So the new parabola meets all the requirements except being discontinuous at $x=0$, it cannot be differentiable there. For $1$, write down the derivative definition for $x=0$. Then instead of letting $\Delta x$ go to $0$, can you find another function which goes to $0$ in some limit? How about $1/n$ as $n$ goes to infinity? So could you replace $\Delta x$ with $1/n$ and let $n$ go to infinity. Try that replacement then use some algebra. Does it look like $1$? – mathematics2x2life Dec 13 '13 at 17:58
@mathematics2x2life : Thanks for 2. As for 1- I am not sure I understand... You only take a specific way to tend to zero ($\frac{1}{n} $ ) ... No one tells you that by taking other ways to tend to zero, you'll get the same result – criticism Dec 14 '13 at 08:43
@criticism See Eric's answer, which was what I was trying to help you see. – mathematics2x2life Dec 14 '13 at 21:34

score 1 · Answer 3 · answered Dec 14 '13 at 20:21

For 1, if $\lim_{n\to\infty}n(f(1/n)-f(0))=0$ then we have $\displaystyle\lim_{n\to\infty}\frac{f(\frac{1}{n})-f(0)}{\frac{1}{n}}=0$. It is common to assume that $n$ denotes a natural number but this was not indicated in the problem statement. So assuming the stronger statement (i.e. that the limit is taken for $n\in\mathbb{R}$) this is equivalent to $$\lim_{h\to 0^{+}}\frac{f(h)-f(0)}{h}=0.$$ This does not mean that $\lim_{h\to 0}\frac{f(h)-f(0)}{h}$ exists. For a counterexample think up a piecewise function. Something like the following: $$f(x)=\left\{\begin{array}{ll} 0& :x\geq0\\h(x)&:x<0 \end{array}\right..$$ Notice that this will satisfy the conditions of 1 regardless of what $h(x)$ is.

Derivative Counterexamples - Calculus

3 Answers3