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**Fistfiz****Member**- Registered: 2012-07-20
- Posts: 33

Hi guys, i'm trying to figure out if this is true or not, can you help me?

Conjecture: Let f:[m,+∞)->R be a continuous and monotonous function with a horizontal asymptote y0 (as x->+∞). Then:

1) f is derivable.

2) f'->0 as x->∞.

I ask for f being monotonous because the only counterexamples, to the non-improved conjecture, that came to my mind are things like f(x)=sin(x^2)/x.

Thanks in advance.

30+2=28 (Mom's identity)

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**scientia****Banned**- Registered: 2009-11-13
- Posts: 224

Not true. may only be piecewise differentiable, not wholly differentiable.

Let us define a sequence of functions as follows:

where

is chosen so that for continuity. Then if we make by piecing together the functions and their domains, this function satisfies all your given conditions but is not differentiable at any integer value of , where the gradient changes abruptly.*Last edited by scientia (2013-01-16 11:11:58)*

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**Fistfiz****Member**- Registered: 2012-07-20
- Posts: 33

uhmm i'm trying to figure out how do you see that f goes to a y0... let's see:

so the c(n) part converges and the x part of course does. Is there a easiest way to see that?

30+2=28 (Mom's identity)

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**Fistfiz****Member**- Registered: 2012-07-20
- Posts: 33

However, we observe that the derivative (where it does exist) goes to zero... maybe moving (1) to the hypothesis... Anyway, that was a good example, thank you

30+2=28 (Mom's identity)

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