Limit of x^cos(1/x) as x approaches 0+ | Calculus Homework Solution

[Snen]

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]

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?

No, that doesn't look right at all.

what happens when $\cos(1/x)$ is negative?

 

[Snen]

No, that doesn't look right at all.

what happens when $\cos(1/x)$ is negative?

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+

 

[Ray Vickson]

Well then y would tend to infinity but the limit is x-> 0+

He is not asking about what happens to x; he is asking about what happens to x^cos(1/x).

 

[Snen]

He is not asking about what happens to x; he is asking about what happens to x^cos(1/x).

tends to infinity I guess