# Homework Help: 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.

#### Attached Files:

• ###### wolframalpha-plot-sin-x2overx.gif
File size:
13.6 KB
Views:
167
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.