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Question Number 216144 by klipto last updated on 28/Jan/25

1. Lim_(n→∞) [(1/n^2 )+(2/n^2 )+(3/n^2 )+...+((n+1)/n^2 )]  2. lim_(x→0) (((3sin5x)/x))^((1−cos4x)/x^2 )

1.Limn[1n2+2n2+3n2+...+n+1n2]2.limx0(3sin5xx)1cos4xx2

Answered by Ghisom last updated on 28/Jan/25

lim_(n→∞)  Σ_(k=1) ^(n+1)  (k/n^2 ) =lim_(n→∞)  ((Σ_(k=1) ^(n+1)  k)/n^2 ) =lim_(n→∞)  ((n^2 +3n+2)/(2n^2 )) =(1/2)

limnn+1k=1kn2=limnn+1k=1kn2=limnn2+3n+22n2=12

Answered by mr W last updated on 28/Jan/25

1)  lim_(n→∞) Σ_(k=1) ^(n+1) (k/n^2 )  =∫_0 ^1 xdx  =(1/2)

1)limnn+1k=1kn2=01xdx=12

Commented by klipto last updated on 28/Jan/25

thanks bossman

thanksbossman

Answered by mr W last updated on 28/Jan/25

2)  =lim_(x→0) (((15 sin 5x)/(5x)))^((2 sin^2  2x)/x^2 )   =lim_(x→0) (((15 sin 5x)/(5x)))^(8(((sin 2x)/(2x)))^2 )   =(15×1)^(8×1^2 )   =15^8

2)=limx0(15sin5x5x)2sin22xx2=limx0(15sin5x5x)8(sin2x2x)2=(15×1)8×12=158

Commented by klipto last updated on 28/Jan/25

thanks ghisom

thanksghisom

Commented by Ghisom last updated on 28/Jan/25

yes. I missed the power...

yes.Imissedthepower...

Answered by MrGaster last updated on 29/Jan/25

(1):=lim_(n→∞) [((1+2+3+…+(n+1))/n^2 )]  =lim_(n→∞) [(((n+1)(n+2))/(2/n^2 ))]  =lim_(n→∞) [((n^2 +3n+2)/(2n^2 ))]  =(1/2)  (2):=lim_(x→0) (3∙((sin 5x)/(5x))∙5)^((1−(1−(((4x^2 ))/(2!))+…))/x^2 )   =lim_(x→0) (15∙((sin 5x)/(5x)))^((8x^2 )/x^2 )   =lim_(x→0) 15^8 (lim_(x→0) ((sin 5x)/(5x))=1)  =15^8

(1):=limn[1+2+3++(n+1)n2]=limn[(n+1)(n+2)2n2]=limn[n2+3n+22n2]=12(2):=limx0(3sin5x5x5)1(1(4x2)2!+)x2=limx0(15sin5x5x)8x2x2=lim15x08(limx0sin5x5x=1)=158

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