Question Number 114981 by bobhans last updated on 22/Sep/20

Without L′Hopital   (1)lim_(x→1)  ((x(x+(1/x))^5 −32)/(x−1)) =?  (2) lim_(x→∞)  (√(2x+(√(2x+(√(2x+(√(2x+(√(...)))))))))) −(√(2x)) = ?

Commented byDwaipayan Shikari last updated on 22/Sep/20

lim_(x→1) ((x(x+(1/x))^5 −2^5 )/( (x)^(1/5) (x+(1/x))−2)).((x^(6/5) +x^(−(4/5)) −2)/(x−1))  lim_(x→1) 5.2^4 .((2/5))=32       (((x^(6/5) +x^(−(4/5)) −2)/(x−1))=(2/5))  ((a^6 +a^(−4) −2)/(a^5 −1)).((a−1)/(a−1))   (x=a^5 )  ((a^(10) +1−2a^4 )/(a^4 +a^3 +a^2 +a+1)).(1/(a^4 (a−1)))  (1/5)(a^(10) +1−2a^4 ).(1/(a^5 −a^4 ))=(1/5)(a^(10) −a^8 +a^8 −2a^4 +1).(1/(a^4 (a−1)))=(1/5)(a^5 +a^4 +(((a^4 −1)^2 )/(a^4 (a−1))))=(2/5)  [(((a^4 −1)(a^4 −1))/(a−1))=(((a−1)^2 (a+1)^2 (a^4 −1))/(a−1))=0]

Commented bybemath last updated on 23/Sep/20

(1)compare by Hopital  lim_(x→1)  ((x(((x^2 +1)/x))^5 −32)/(x−1)) = lim_(x→1)  (((x^2 +1)^5 −32x^4 )/(x^4 (x−1)))  lim_(x→1)  ((5.2x(x^2 +1)^4 −128x^3 )/(5x^4 −4x^3 )) =  ((10.2^4 −128)/1) = 160−128=32

Answered by john santu last updated on 22/Sep/20

(2) let (√(2x+(√(2x+(√(2x+(√(2x+(√(...)))))))))) = b  ⇔ 2x +b = b^2  ⇒2x=b^2 −b  lim_(b→∞) b−(√(b^2 −b)) = lim_(b→∞) ((b^2 −(b^2 −b))/(b+(√(b^2 −b))))  = lim_(b→∞) (b/(b(1+(√(1−(1/b)))))) = (1/2)

Answered by Dwaipayan Shikari last updated on 22/Sep/20

(√(2x+(√(2x+(√(2x+...)))))) =p  2x+p=p^2   p^2 −p−2x=0  p=((1±(√(1+8x)))/2)=((1+(√(1+8x)))/2)  (1/2)(1+(√(1+8x))−2(√(2x)))=(1/2)+((1+8x−8x)/( (√(1+8x))+2x))=(1/2)