Question about a function

  • #1
391
0
can exist an smooth function with the property

[tex] y(\infty) =0 [/tex] and [tex] y'(\infty) =1 [/tex] ?

the inverse case, a function that tends to 1 for big x and whose derivative tends to 0 is quite obvious but this case i am not sure if there will exist
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
4,302
47
I suppose that you mean
[tex]\lim_{x \to \infty} y(x) = 0[/tex] and [tex]\lim_{x \to \infty} y'(x) = 1[/tex] ?

Actually, I think that for a smooth function to have a limit at infinity, the derivative should have limit 0 (at least that's what my intuition tells me: for the function to have a limit at infinity, it should become progressively more flat, so it doesn't run away from its limit value).

I have some other work now, but I will try to prove that rigorously later (if you want, give it a try yourself).
 
  • #3
133
0
It follows easily from MVT.
 
  • #4
CompuChip
Science Advisor
Homework Helper
4,302
47
That's what I figured, but I got caught up in epsilons and deltas on the back of my scrap piece of paper.
After finishing my Saturday's to-do list I will take a completely blank paper of normal size and try again :)
 
  • #5
133
0
On the second thought, derivative doesn't have to go to zero, unfortunetely. Consider [tex]\frac{\sin (x^2)}{x}[/tex]. It clearly tends to zero, yet the derivative oscilates. Still, the derivative cannot tend to a nonzero number, and this follows from MVT for sure :wink:
Sorry for the mistake.

Edit: derivative of [tex]\frac{\sin (x^2)}{\sqrt{x}}[/tex] oscilates unboundedly, while the function goes to zero.
 
Last edited:

Related Threads on Question about a function

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
2K
Replies
2
Views
1K
Replies
1
Views
2K
Replies
5
Views
4K
Replies
4
Views
2K
Replies
4
Views
2K
Replies
1
Views
2K
Replies
9
Views
747
Top