Understanding Limits at Infinity and Non-Existence of Derivatives

[matts0]

Homework Statement

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;
lim_x->∞f(x) =2;
lim_x->∞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?
Thanks in advance.

Homework Equations

The Attempt at a Solution

 

[LCKurtz]

Homework Statement

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;
lim_x->∞f(x) =2;
lim_x->∞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?
Thanks in advance.

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.

 

[SammyS]

Welcome to the PH Forums !

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

It says lim_x➙∞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.

 

[matts0]

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

 

[SammyS]

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

 

[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.

 

[SammyS]

Yes, that function works, but your derivative has a couple of errors in it.

Click on the http://www.wolframalpha.com/input/?i=2+sin(x3)/x" and look down the page for the derivative.

 

[matts0]

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