Asked by Anonymous
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?
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?
Answers
Answered by
Count Iblis
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 = (x-a)(x+9a)
Therefore:
(x^2 + 8ax - 9a^2)/(x - a) = (x+9a)
Lim x ---> a of (x+9a) = 10 a
= 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 = (x-a)(x+9a)
Therefore:
(x^2 + 8ax - 9a^2)/(x - a) = (x+9a)
Lim x ---> a of (x+9a) = 10 a
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