How does #\lim_(n\rarr\infty)(2|x-2|)/(n+3)=0#?

Question

Without doing any further work, is it possible to recognize the limit as 0?
Or should I simplify further (if so, then how?)

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Answer 1 · 2018-04-26T17:51:12

show below

show the steps and focus in the explanition

since #nrarroo# n is a variable value and others like: x are constant

x=constant

any constant divide by#oo#equal zero

c=any constant
#c/oo=0#

#c/-oo=0#

in addition any constant plus #oo# or #-oo# equal
#c+(+oo)=oo# and
#c+(-oo)=-oo#

#lim_(nrarroo)[(2|x-2|)/(n+3)]=(2|x-2|)/(oo)=0#

#(2|x-2|)=#constant

#lim_(nrarroo)[(2|x-2|)/(n+3)]=0#