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

1 Answer
Feb 20, 2018

#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]}