# Lim x->0+ (x^cos(1/x))

## 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
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## 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?

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
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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).

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