The discussion revolves around evaluating the limit of the expression x^cos(1/x) as x approaches 0 from the positive side, within the context of calculus.
Some participants express skepticism about the correctness of the initial approach, particularly regarding the implications of cos(1/x) being negative. There is an ongoing exploration of how this affects the limit and the behavior of the expression.
Participants are considering the implications of the oscillatory nature of cos(1/x) as x approaches 0, which introduces complexity into the limit evaluation.
No, that doesn't look right at all.Snen said:Homework Statement
Homework Equations
The Attempt at a Solution
let y = lim x->0+ x^cos(1/x)
lny = cos(1/x)*lnx = (x*cos(1/x)) * (lnx/x)
x*cos(1/x) = 0 (sandwich theorem)
lnx/x = 0 (l'hopital)
so lny = 0
and y = 1
Is this correct?
PeroK said:No, that doesn't look right at all.
what happens when ##\cos(1/x)## is negative?
Well then y would tend to infinity but the limit is x-> 0+PeroK said:No, that doesn't look right at all.
what happens when ##\cos(1/x)## is negative?
Snen said:Well then y would tend to infinity but the limit is x-> 0+
tends to infinity I guessRay Vickson said:He is not asking about what happens to x; he is asking about what happens to x^cos(1/x).