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Question Number 109763 by Karani last updated on 25/Aug/20

lim_(n→∞) ((3/(1−^n (√8)))−(5/(1−^n (√(32)))))=?

limn→∞(31−n8−51−n32)=?

Answered by bemath last updated on 25/Aug/20

▽((♭e)/(math))Δ lim_(n→∞) ((3−3 ((32))^(1/(n )) −5+5 (8)^(1/n) )/(1−((32))^(1/n) −(8)^(1/n) +((256))^(1/(n )) )) = let (1/n) = t { ((n→∞)),((t→0)) :} lim_(t→0) [((−2−3 (32)^t +5 (8)^t )/(1−(32)^t −(8)^t +(256)^t )) ]= lim_(t→0) [((−2−3(2^t )^5 +5(2^t )^3 )/(1−(2^t )^5 −(2^t )^3 +(2^t )^8 )) ]= set 2^t = m,m→1 lim_(m→1) [((5m^3 −3m^5 −2)/(m^8 −m^3 −m^5 +1))]= L′Hopital lim_(m→1) [((15m^2 −15m^4 )/(8m^7 −3m^2 −5m^4 )) ]= lim_(m→1) [((15−15m^2 )/(8m^5 −3−5m^2 )) ] = lim_(m→1) [((−30m)/(40m^4 −10m)) ] = ((−30)/(30))=−1

▽♭emathΔlimn→∞3−332n−5+58n1−32n−8n+256n=let1n=t{n→∞t→0limt→0[−2−3(32)t+5(8)t1−(32)t−(8)t+(256)t]=limt→0[−2−3(2t)5+5(2t)31−(2t)5−(2t)3+(2t)8]=set2t=m,m→1limm→1[5m3−3m5−2m8−m3−m5+1]=L′Hopitallimm→1[15m2−15m48m7−3m2−5m4]=limm→1[15−15m28m5−3−5m2]=limm→1[−30m40m4−10m]=−3030=−1

Answered by Her_Majesty last updated on 25/Aug/20

let n=(1/t) with t→0^+ (3/(1−8^t ))−(5/(1−32^t ))=−((3×8^t +6×4^t +4×2^t +2)/((4^t +2^t +1)(16^t +8^t +4^t +2^t +1))) t=0 −((15)/(3×5))=−1

letn=1twitht→0+31−8t−51−32t=−3×8t+6×4t+4×2t+2(4t+2t+1)(16t+8t+4t+2t+1)t=0−153×5=−1

Answered by john santu last updated on 25/Aug/20

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