How do you find #lim root3(t+4)-root3t# as #t->oo#?

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Answer 1 · 2017-03-01T18:35:44

#0#

#root3(t+4)-root3t = root3(t)(root3(1+4/t)-1)#

but

#root3(1+4/t) = 1+1/3(4/t)+(1/3(1/3-1))/(2!)(4/t)^2+cdots#

and

#root3(t)(root3(1+4/t)-1) = root3(t)(-1/3(4/t)-(1/3(1/3-1))/(2!)(4/t)^2-cdots) =#

#=1/t^(2/3)(-4/3+f(1/t))# so

#lim_(t->oo)root3(t+4)-root3t = lim_(t->oo)1/t^(2/3)(-4/3+f(1/t))=0#