<p>lim
x→1
(5x + 1)1/3 −(x + 5)1/3
(2x + 3)1/2 −(x + 4)1/2 =
m
√
5
n(2n)2/3 ,
where gcd(m, n) = 1. Then 8m + 12n is equal to</p>
Step-by-Step Solution
Key Concept: When both numerator and denominator vanish at the same point, use first-order linearization.
<p>At x = 1, both numerator and denominator are 0. So use derivatives at x = 1.</p> For the numerator, d dx(5x + 1)1/3 = 5 3(5x + 1)-2/3, d dx(x + 5)1/3 = 1 3(x + 5)-2/3. Hence at x = 1, N′(1) = 5 3 \cdot 6-2/3 -1 3 \cdot 6-2/3 = 4 3 \cdot 6-2/3. For the denominator, d dx(2x + 3)1/2 = 1 \sqrt{2x} + 3, d dx(x + 4)1/2 = 1 2\sqrt{x} + 4. Thus, D′(1) = 1 \sqrt 5 - 1 2 \sqrt 5 = 1 2 \sqrt<p><strong>5</strong>: </p> Therefore, lim x\to 1 N(x) D(x) = N′(1) D′(1) = 4 36-2/3 1 2 \sqrt 5 = 8 \sqrt 5 3 \cdot 62/3 . This is of the form m \sqrt 5 n(2n)2/3 with m = 8 and n = 3. Hence, 8m + 12n = 8 \cdot 8 + 12 \cdot 3 = 64 + 36 = 100. Shortcut / Fast View For u(x)p -v(x)p near a common point, replace each by its tangent-line approximation.
Correct Answer: (100)