Question Number 11444 by Joel576 last updated on 26/Mar/17

If lim_(x→0) (((√(px + q)) − 2)/x) = 1 What is the value of p + q ?

Answered by ajfour last updated on 26/Mar/17

then lim_(x→0) (((√q)(1+px/q)^(1/2) −2)/x) =1 ⇒ lim_(x→0) (((√q)(1+((px)/(2q)) ) −2)/x) =1 = lim_(x→0) ((((√q)−2)+(((px)/(2(√q)))))/x) =1 ⇒ (√q)=2 or q=4 then (p/(2(√q))) =1 as (√q) =2 , p=4 p+q=8

Commented byJoel576 last updated on 26/Mar/17

thank you very much

Answered by ridwan balatif last updated on 26/Mar/17

lim_(x→0) (((√(px+q))−2)/x)=1 test limit: (((√(p×0+q))−2)/0)=1→this is Impossible so form of the test limit should be (0/0) (√(p×0+q))−2=0 (√q)−2=0 q=4 lim_(x→0) (((√(px+4))−2)/x)=1 lim_(x→0) (((((√(px+4))−2))/x)×((((√(px+4))+2)/((√(px+4))+2))))=1 lim_(x→0) (((px+4−4)/(x((√(px+4))+2)))=1 lim_(x→0) ((px)/(x((√(px+4))+2)))=1 lim_(x→0) (p/((√(px+4))+2))=1 (p/((√(p×0+4))+2))=1 (p/(2+2))=1 p=4 q=4 ∴p+q=8