Let f(b)= ((2)/(b-9)) - ((36)/(b^2-81))?

Question

calculate lim-->9 f(b) by first finding a continuous function which is equal to f everywhere except b=9

Answer 1

separate the limits

lim f(b) = lim 2/(b-9) - lim 36/(b^2-81)
multiply the first limit ny (b+9)/(b+9)

lim f(B)= lim 2(b+9)/(b^2-81) - lim 36/(b^2-81)
combine two limits

lim f(b) = (2b + 18 -36) / (b+9)(b-9)
lim f(b) = 2(b-9) / (b+9)(b-9)
(b-9) cancels out

lim f(b) = 2 / (b+9)
as b approaches 9

lim f(b) = 2 / 18 = 1/9 ------> answer

Answer 2

Use b^2 - 81 = (b+9)(b-9) as common denominator to get
f(b) = (2(b+9) - 36)/((b+9)(b-9), then the b-9 factor will cancel and you can evaluate directly. The continuous function they are asking for is the one you get after canceling -- it's the same as the original except it has a well defined value at b = 9.

Answer 3

(b + 9)(b - 9) = b^2 - 81

let b=9+h

2/(b - 9) - 36/(b^2 - 81) =
= 2/(b - 9) - 36/[(b - 9)(b + 9)] =
= 2/h - 36/[h(h + 18)] = 2(h + 18)/[h(h + 18)] - 36/[h(h - 18)] =
= [2(h + 18) - 36]/[h(h + 18)] =
= (2h + 36 - 36)/[h(h + 18)] = 2h/[h(h + 18)]

lim (b -> 9) f(b) = lim(h -> 0) 2h/[h(h + 18)] =
= lim(h -> 0) 2/(h + 18) = 2/18 = 1/9

Answer 4

-2/9-36/(b^2-81)