How do you find the limit of #(sin^4x)/(x^(1/2))# as x approaches infinity?

Question

1720 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-14T08:53:25

#0#

#sin^4 x in [-1, 1]#, #x in mathcal(R)#.

Because the numerator is continuous and bounded:

#lim_(x to oo) (sin^4x)/(x^(1/2))#

#= sin^4x lim_(x to oo)1/(x^(1/2))#

by quotient rule
#= sin^4x(lim_(x to oo) 1)/( lim_(x to oo) x^(1/2))#

#= sin^4x * 1/( oo)#

#=0#