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

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