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Question Number 193262 by josemate19 last updated on 09/Jun/23

lim_(x→+∞) (ln((x+(√(x^2 +1)))/(x+(√(x^2 −1)))).ln^2 ((x+1)/(x−1)))

limx→+∞(lnx+x2+1x+x2−1.ln2x+1x−1)

Answered by qaz last updated on 09/Jun/23

lim_(x→+∞) ((x+(√(x^2 +1)))/(x+(√(x^2 −1))))=lim_(x→+∞) ((x+1)/(x−1))=1 ⇒lim_(x→+∞) ln((x+(√(x^2 +1)))/(x+(√(x^2 −1))))ln^2 ((x+1)/(x−1))=0

limx→+∞x+x2+1x+x2−1=limx→+∞x+1x−1=1⇒limlnx→+∞x+x2+1x+x2−1ln2x+1x−1=0

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