- #1

- 3

- 0

## 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?

- #1

- 3

- 0

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?

- #2

- 13,696

- 6,187

No, that doesn't look right at all.## 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?

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

- #3

- 3

- 0

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+No, that doesn't look right at all.

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

- #4

Ray Vickson

Science Advisor

Homework Helper

Dearly Missed

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He is not asking about what happens to x; he is asking about what happens to x^cos(1/x).Well then y would tend to infinity but the limit is x-> 0+

- #5

- 3

- 0

tends to infinity I guessHe is not asking about what happens to x; he is asking about what happens to x^cos(1/x).

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