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Question Number 117545 by Lordose last updated on 12/Oct/20

lim_(x→0^+ ) (1+tan^2 ((√x)))^(1/(2x))

limx0+(1+tan2(x))12x

Answered by TANMAY PANACEA last updated on 12/Oct/20

lny=lim_(x→0+)  ((ln(1+tan^2 ((√(x))) ))/(tan^2 ((√x) )))×((tan((√x) ))/( (√x)))×((tan((√x) ))/( (√x)))×(1/2)  lny=(1/2)→y=e^(1/2)

lny=limx0+ln(1+tan2(x))tan2(x)×tan(x)x×tan(x)x×12lny=12y=e12

Answered by Olaf last updated on 12/Oct/20

(1/(2x))ln(1+tan^2 ((√x))) ∼_0  (1/(2x))ln(1+((√x))^2 ) ∼_0  (1/(2x))×x = (1/2)  ln(1+tan^2 ((√x)))^(1/(2x))  ∼_0  (1/2)  ⇒ lim_(x→0^+ ) (1+tan^2 ((√x)))^(1/(2x))  = e^(1/2)

12xln(1+tan2(x))012xln(1+(x)2)012x×x=12ln(1+tan2(x))12x012limx0+(1+tan2(x))12x=e12

Answered by Dwaipayan Shikari last updated on 12/Oct/20

lim_(x→0^+ ) (1+tan^2 (√x))^(1/(2x)) =lim_(x→0^+ ) (1+((√x))^2 )^(1/(2x)) =((1+x)^(1/x) )^(1/2) =(√e)  tan(√x)→(√x)

limx0+(1+tan2x)12x=limx0+(1+(x)2)12x=((1+x)1x)12=etanxx

Answered by bobhans last updated on 12/Oct/20

L = lim_(x→0^+  )  (1+tan^2 ((√x)))^(1/(2x))   L = e^(lim_(x→0^+ ) (1+tan^2 ((√x))−1)×(1/(2x)))   L= e^(lim_(x→0^+ ) (((tan^2 ((√x)))/(2((√x))^2 ))))  = e^((1/2)×(lim_(x→0^+ ) ((tan (√x))/( (√x))))^2 )   L = e^(1/2)  = (√e) .

L=limx0+(1+tan2(x))12xL=elimx0+(1+tan2(x)1)×12xL=elimx0+(tan2(x)2(x)2)=e12×(limx0+tanxx)2L=e12=e.

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