The solution goes like this
From the given #lim_(x rarr oo) x^sin (1/x)#
We test first. #lim_(x rarr oo) x^sin (1/x)# this results to #oo^0#
We apply logarithms
Let #y=x^sin (1/x)#
Take the logarithm of both sides
#y=x^sin (1/x)#
#ln y=ln x^sin (1/x)#
#ln y=sin (1/x)*ln x#
#ln y=sin (1/x)*ln x#
Test the limit of #ln y# and it takes the form #0*oo#
We arrange it so that it will be #oo/oo#
#lim_(x rarr oo) ln y=lim_(x rarr oo) (ln x)/(1/sin (1/x))=oo/oo#
Use continuous derivatives on both numerator and denominator until we come up with a real number.
#lim_(x rarr oo) ln y=lim_(x rarr oo) (1/x)/((sin (1/x)*0-1*cos(1/x)(-1/x^2))/(sin^2 (1/x)))#
#lim_(x rarr oo) ln y=lim_(x rarr oo) (1/x)/((1*cos(1/x)(1/x^2))/(sin^2 (1/x)))=lim_(x rarr oo) (sin^2 (1/x))/(1/x*cos(1/x))=0/(0*1)=0/0#
#lim_(x rarr oo) ln y=0/0#
Differentiate again
#lim_(x rarr oo) ln y=#
#=lim_(x rarr oo) (2sin (1/x)cos (1/x)*(-1/x^2))/(1/x(-sin (1/x)(-1/x^2))+cos (1/x)(-1/x^2))#
#=lim_(x rarr oo) (2sin (1/x)cos (1/x))/(-1/x*sin (1/x)+cos (1/x))=(2*0*1)/(0+1)=0/1=0#
We now have
#lim_(x rarr oo) ln y=0#
Finally,
#lim_(x rarr oo) y=e^0=1#
And therefore
#color(red)(lim_(x rarr oo) x^sin (1/x)=1)#
God bless....I hope the explanation is useful.
