Find the limit #lim_"n→ oo" cosn^3/(2n)- (3n)/(6n+1)#?

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#lim_"n→ oo" cosn^3/(2n)- (3n)/(6n+1)#

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Answer 1 · 2018-03-13T13:26:55

#lim_(n->oo) cos(n^3)/(2n) - (3n)/(6n+1) =-1/2#

The limits of both addends are finite, and we can evaluate them separately:

#lim_(n->oo) cos(n^3)/(2n) = 0#

as #abs(cos(n^3)) <=1#, so the numerator is bounded and the denominator tends to #+oo#.

#lim_(n->oo) (3n)/(6n+1) = lim_(n->oo) 3/(6+1/n) = 3/6 = 1/2#

Then:

#lim_(n->oo) cos(n^3)/(2n) - (3n)/(6n+1) = 0-1/2 = -1/2#