Find the limit of the sequence #ln(n^2+2) -1/2ln(n^4+4)#. Does it converge to 0?

1 Answer
George C.
Mar 17, 2017

#lim_(n->oo) (ln(n^2+2)-1/2ln(n^4+4)) = 0#

Explanation:

#ln(n^2+2)-1/2ln(n^4+4) = 1/2ln((n^2+2)^2)-1/2ln(n^4+4)#

#color(white)(ln(n^2+2)-1/2ln(n^4+4)) = 1/2(ln(n^4+4n^2+4)-ln(n^4+4))#

#color(white)(ln(n^2+2)-1/2ln(n^4+4)) = 1/2ln((n^4+4n^2+4)/(n^4+4))#

#color(white)(ln(n^2+2)-1/2ln(n^4+4)) = 1/2ln(1+(4n^2)/(n^4+4))#

#color(white)(ln(n^2+2)-1/2ln(n^4+4)) = 1/2ln(1+4/(n^2+4/n^2))#

Hence:

#lim_(n->oo) (ln(n^2+2)-1/2ln(n^4+4)) = lim_(n->oo) 1/2ln(1+4/(n^2+4/n^2))#

#color(white)(lim_(n->oo) (ln(n^2+2)-1/2ln(n^4+4))) = 1/2ln(1+0)#

#color(white)(lim_(n->oo) (ln(n^2+2)-1/2ln(n^4+4))) = 0#

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