Posted by Anonymous on Tuesday, April 15, 2008 at 8:29pm.
Evaluate limit, x > a, [(x + 4a)^2  25a^2] / [x  a]
My work:
= limit, x > a, (x^2 + 8ax + 16a^2  25a^2) / (x  a)
= limit, x > a, (x^2 + 8ax  9a^2) / (x  a)
= limit, x > a, (x + 8a  9a^2) / (a)
= (a + 8a  9a^2) / (a)
= 9a^2  8a + 1
Textbook Answer: 10a
What did I do wrong, please correct it?

Calculus; Limits  Count Iblis, Tuesday, April 15, 2008 at 9:53pm
My work:
= limit, x > a, (x^2 + 8ax + 16a^2  25a^2) / (x  a)
= limit, x > a, (x^2 + 8ax  9a^2)/
(x  a)
x^2 + 8ax  9a^2 = (xa)(x+9a)
Therefore:
(x^2 + 8ax  9a^2)/(x  a) = (x+9a)
Lim x > a of (x+9a) = 10 a