How do you find the limit #(x+x^-2)/(2x+x^-2)# as #x->oo#?

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Answer 1 · 2018-03-08T08:16:23

The limit is #=1/2#

Let's rewrite the function

#y=(x+x^-2)/(2x+x^-2)=(x+1/x^2)/(2x+(1/x^2))#

#=(x^3+1)/(2x^3+1)#

#=(x^3(1+1/x^3))/(x^3(2+1/x^3))#

#=(1+1/x^3)/(2+1/x^3)#

#Lim_(x->oo)(1/x^3)=0 #

Therefore,

#Lim_(x->oo)(y)=Lim_(x->oo)((1+1/x^3)/(2+1/x^3))=(1+0)/(2+0)=1/2#