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