Question

How do i find the limit as x approaches to infinity of `x^sin(1/x)` ?

Answer 1

`lim_(x->oo) x^sin(1/x) = 1`

Explanation:

Write the function as:

`x^sin(1/x) = (e^lnx)^sin(1/x) = e^(lnxsin(1/x))`

Consider now the limit:

`lim_(x->oo) lnxsin(1/x)`

if we write it in the form:

`lim_(x->oo) lnxsin(1/x) = lim_(x->oo) sin(1/x)/(1/lnx)`

it is in the indeterminate form `0/0` and we can apply l'Hospital's rule:

`lim_(x->oo) lnxsin(1/x) = lim_(x->oo) (d/dx sin(1/x))/(d/dx (1/lnx))`

`lim_(x->oo) lnxsin(1/x) = lim_(x->oo) (-cos(1/x)/x^2)/(-1/(xln^2x))`

`lim_(x->oo) lnxsin(1/x) = lim_(x->oo) cos(1/x) ln^2x/x`

Using l'Hosiptal's rule again:

`lim_(x->oo) ln^2x/x = lim_(x->oo) (2lnx)/x = lim_(x->oo) 2/x = 0`

and then:

`lim_(x->oo) lnxsin(1/x) = lim_(x->oo) cos(1/x) ln^2x/x = 0`

As `e^x` is continuous for `x=0`:

`lim_(x->oo) e^(lnxsin(1/x)) = e^(( lim_(x->oo) lnxsin(1/x))) = e^0 = 1`

graph{x^sin(1/x) [-10, 10, -5, 5]}