Question Number 114299 by bemath last updated on 18/Sep/20

 solve lim_(p→∞)  (((p^2 +2)/(p+2)))^((p+2)/(p^2 +2))  ?

Commented bymohammad17 last updated on 18/Sep/20

put:k=((p+2)/(p^2 +2))⇒k→0    lim_(k→0) ((1/k))^k ⇒ln(A)=lim_(k→0) ((ln((1/k)))/(1/k))=0  ln(A)=0⇒A=e^0 =1

Answered by 1549442205PVT last updated on 18/Sep/20

I=lim (_(p→∞t) ((p^2 +2)/(p+2)))^((p+2)/(p^2 +2))    lnI=lim_(p→∞) [((p+2)/(p^2 +2))ln(((p^2 +2)/(p+2)))]  =lim_(p→∞) ((ln(((p^2 +2)/(p+2))))/((p^2 +2)/(p+2))).This is the form(∞/∞)  Applying L′Hopital rule we have  lnI=lim_(p→∞) ((((p+2)/(p^2 +2))×((2p(p+2)−(p^2 +2))/((p+2)^2 )))/((2p(p+2)−(p^2 +2))/((p+2)^2 )))  =lim_(p→∞) ((p+2)/(p^2 +2))=0⇒I=e^0 =1

Answered by mathmax by abdo last updated on 19/Sep/20

let f(x) =(((x^2  +2)/(x+2)))^((x+2)/(x^2  +2))  ⇒f(x) =e^(((x+2)/(x^2  +2))ln(((x^2  +2)/(x+2))) )  we have  at +∞   ((x+2)/(x^2  +2))ln(((x^2  +2)/(x+2)))∼(1/x)ln(x)→0 ⇒lim_(x→+∞) f(x)=e^0  =1  =