Question on limit at infinity

1. Sep 26, 2011

matts0

1. The problem statement, all variables and given/known data
Hello everyone, I am just new to this forum and also a beginner at calculus.
I have a question from my textbook. It's:
Find an example of f(x) that satisfies the following conditions :
f(x) is differentiable for all x>0;
limx->∞f(x) =2;
limx->∞f'(x) does not exist;

I think that if f(x) satisfies the second condition it must have a horizontal tangent at infinity,which means f'(x) = 0 at infinity, am I right? and what does "f'(x) does not exist" really mean?
2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 26, 2011

LCKurtz

You are right about the horizontal asymptote (not horizontal "tangent") of y = 2 as x → ∞. And many graphs you have seen to have the curve "leveling out" as the graph approaches the asymptote, in which case you would have

$$\lim_{x\rightarrow \infty}f'(x)=0$$

What you need to do is find an example that has the y = 2 asymptote but the slope doesn't get close to 0, maybe because it "wobbles back and forth", to phrase it informally.

Last edited: Sep 26, 2011
3. Sep 26, 2011

SammyS

Staff Emeritus
Welcome to the PH Forums !!!

It doesn't say f'(x) does not exist.

It says limx➙∞f'(x) does not exist.

If you're just beginning Calculus, this problem could be difficult.

You are correct as far as rational functions are concerned.

Try to modify a sine or cosine function, so that its amplitude decreases as x➙∞ , but oscillates more and more rapidly as x➙∞ . If it oscillates rapidly enough, then the derivative may oscillate with constant or increasing amplitude. Therefore, the limit derivative will not converge as x➙∞ .

See the image of a graph of such a first derivative, f'(x). ...Plotted in WolframAlpha.

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4. Sep 26, 2011

matts0

Actually I confused "tangent" with "asymptote"... Thanks for the help from both of you.

5. Sep 26, 2011

SammyS

Staff Emeritus
So, what did you come up with for f(x) ?

6. Sep 27, 2011

matts0

I came up with this :
f(x)=2+sin(x3)/x
so f'(x) = sin(x3)/x2-3cos(x3) and it is not stable at infinity because of cos(x3),right?
Please tell me if it is ok or not. Thank you.

7. Sep 27, 2011

SammyS

Staff Emeritus
Last edited by a moderator: Apr 26, 2017
8. Sep 27, 2011

matts0

Thank you.I should have done more exercises.
and also thanks for introducing WolframAlpha to me. That's really helpful.