Question

here is a tough one. My professor and I are arguing on the existence of a function, we know it exists, but are having a difficult time proving its existence and exampling it. the function has two requirements: lim(t->infinity)f(x)=c and lim(t->infinity)f'(x)≠0. preferably the function is a non trigonometric one, but other trancendentals are fine.

Answer 1

Hang on, lim as t approaches infinity? Is x to be treated as a constant, then?

Answer 2

typo, x is t

Answer 3

What if the limit of the derivative was infinity (does not exist), does that qualify?

Answer 4

possibly. we are looking for if the derivative does not exist (fluctuations), but perhaps infinity will work

Answer 5

Try
\[f(t) = (1 + \frac{1}{t})^{t}\]

Answer 6

hmm the defining equation of e... i would have never considered that, lemmy see

Answer 7

Scratch that, it doesn't work :(

Answer 8

? the derivative does exist lol

Answer 9

I never said it didn't, but upon correcting some mistakes on my part, the limit does go to 0 as well, so...

Answer 10

ahh kk, well im going to try to modify it and see what can work

Answer 11

ok yeah that doesnt work :/

Answer 12

How are you sure this function even exists?

Answer 13

its a well known mathematic issue in advanced diffy q problems. the function has been written and the most common case is a wiggle function but exampling a wiggle function that approaches a given value c is the difficulty

Answer 14

How about f(x)= (sin x^2)/x ?

As x goes to infinity, f goes to zero, but the limit of f'(x) = 2 cos(x^2) - sin(x^2)/x^2 doesn't exist.

Answer 15

The derivative goes to 1

Answer 16

Errrr 3, not 1

Answer 17

Actually, sin(x^2)/x^2 goes to zero as x goes to infinity, and cos(x^2) oscillates between +-1 so it has no limit. The limit as x goes to zero is one.

Answer 18

Ohno waiti misread the derivative. Ill check it later,i have class rightnow