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Question Number 46268 by Saorey last updated on 23/Oct/18
please help me! L=lim_(x→1) ((p/(1−x^p ))−(q/(1−x^q ))) , (p,q∈R)
pleasehelpme!L=limx→1(p1−xp−q1−xq),(p,q∈R)
Commented by maxmathsup by imad last updated on 23/Oct/18
let A(x)=(p/(1−x^p )) −(q/(1−x^q )) changement x=1+t give A(x)=B(t) ⇒lim_(x→1) A(x)=lim_(t→0) B(t) with B(t)=(p/(1−(1+t)^p )) −(q/(1−(1+t)^q )) but (1+t)^p ∼1+pt +((p(p−1))/2)t^2 and (1+t)^q ∼1+qt +((q(q−1))/t) t^2 (t →o) ⇒ B(t)∼(p/(−pt−((p(p−1))/2)t^2 )) −(q/(−qt−((q(q−1))/2)t^2 )) ⇒ B(t) ∼ −(1/(t +((p−1)/2)t^2 )) +(1/(t +(((q−1))/2)t^2 )) =((−t −(((q−1))/2)t^2 +t +((p−1)/2) t^2 )/(t^2 (1+(((p−1)t)/2))(1+((q−1)/2)t))) =(((p−q)/2)/((1+((p−1)/2)t)(1+((q−1)/2)t))) →((p−q)/2) (t→0) ⇒L=lim_(x→1) A(x)=((p−1)/2) .
letA(x)=p1−xp−q1−xqchangementx=1+tgiveA(x)=B(t)⇒limx→1A(x)=limt→0B(t)withB(t)=p1−(1+t)p−q1−(1+t)qbut(1+t)p∼1+pt+p(p−1)2t2and(1+t)q∼1+qt+q(q−1)tt2(t→o)⇒B(t)∼p−pt−p(p−1)2t2−q−qt−q(q−1)2t2⇒B(t)∼−1t+p−12t2+1t+(q−1)2t2=−t−(q−1)2t2+t+p−12t2t2(1+(p−1)t2)(1+q−12t)=p−q2(1+p−12t)(1+q−12t)→p−q2(t→0)⇒L=limx→1A(x)=p−12.
Commented by maxmathsup by imad last updated on 23/Oct/18
L=lim_(x→1) A(x)=((p−q)/2) .
L=limx→1A(x)=p−q2.
Answered by MJS last updated on 23/Oct/18
lim_(x→1) (p/(1−x^p ))−(q/(1−x^q ))=lim_(x→1) ((p(1−x^q )−q(1−x^p ))/((1−x^p )(1−x^q )))= [l′Hopital, 2 times because 1^(st) derivate is still undefined] =lim_(x→1) (((d^2 /dx^2 )[p(1−x^q )−q(1−x^p )])/((d^2 /dx^2 )[(1−x^p )(1−x^q )]))= =lim_(x→1) ((pq(p−1)x^(p−2) +pq(1−q)x^(q−2) )/((p+q)(p+q−1)x^(p+q−2) −p(p−1)x^(p−2) −q(q−1)x^(q−2) ))= =((p−q)/2)
limx→1p1−xp−q1−xq=limx→1p(1−xq)−q(1−xp)(1−xp)(1−xq)=[l′Hopital,2timesbecause1stderivateisstillundefined]=limx→1d2dx2[p(1−xq)−q(1−xp)]d2dx2[(1−xp)(1−xq)]==limx→1pq(p−1)xp−2+pq(1−q)xq−2(p+q)(p+q−1)xp+q−2−p(p−1)xp−2−q(q−1)xq−2==p−q2
Commented by Saorey last updated on 23/Oct/18
can you calculate it without lopital?
canyoucalculateitwithoutlopital?
Answered by ajfour last updated on 24/Oct/18
L=lim_(x→1) {(p/(1−[1−(1−x)]^p ))−(q/(1−[1−(1−x)]^q ))} = lim_(h→0) {(p/(1−(1−ph+((p(p−1)h^2 )/2)−..)))−(q/(1−(1−qh+((q(q−1)h^2 )/2)−..)))} ⇒ L = lim_(h→0) {(1/h)[(1/(1−(((p−1)h)/2)))−(1/(1−(((q−1)h)/2)))]} = lim_(h→0) (1/h)[(((1−(((q−1)h)/2))−(1−(((p−1)h)/2)))/((1−(((p−1)h)/2))(1−(((q−1)h)/2))))] =(((((p−q)/2)))/1) =((p−q)/2) .
L=limx→1{p1−[1−(1−x)]p−q1−[1−(1−x)]q}=limh→0{p1−(1−ph+p(p−1)h22−..)−q1−(1−qh+q(q−1)h22−..)}⇒L=limh→0{1h[11−(p−1)h2−11−(q−1)h2]}=limh→01h[(1−(q−1)h2)−(1−(p−1)h2)(1−(p−1)h2)(1−(q−1)h2)]=(p−q2)1=p−q2.

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