If $$\lim_{x\to 1}\frac{(5x+1)^{1/3}-(x+5)^{1/3}}{(2x+3)^{1/2}-(x+4)^{1/2}} = \frac{m\sqrt{5}}{n(2n)^{2/3}}$$, where gcd(m, n) = 1, then $$8m + 12n$$ is equal to ______.

Using L'Hôpital or Taylor series for the limit as x→1:

Let f(x) = (5x+1)^(1/3) - (x+5)^(1/3) and g(x) = (2x+3)^(1/2) - (x+4)^(1/2)

f(1) = 6^(1/3) - 6^(1/3) = 0, g(1) = 5^(1/2) - 5^(1/2) = 0

f'(x) = 5/(3(5x+1)^(2/3)) - 1/(3(x+5)^(2/3))

f'(1) = 5/(3·6^(2/3)) - 1/(3·6^(2/3)) = 4/(3·6^(2/3))

g'(x) = 1/√(2x+3) - 1/(2√(x+4))

g'(1) = 1/√5 - 1/(2√5) = 1/(2√5)

Limit = f'(1)/g'(1) = [4/(3·6^(2/3))]·[2√5/1] = 8√5/(3·6^(2/3))

= 8√5/(3·(2·3)^(2/3)) = 8√5/(3·2^(2/3)·3^(2/3)) = 8√5/(3^(5/3)·2^(2/3))

Comparing with m√5/(n·(2n)^(2/3)): we need n such that n·(2n)^(2/3) = 3^(5/3)·2^(2/3)

Try n=3: 3·6^(2/3) = 3·6^(2/3) = 3^(5/3)·2^(2/3) ✓ (since 3·(2·3)^(2/3) = 3·2^(2/3)·3^(2/3) = 3^(5/3)·2^(2/3))

So m = 8, n = 3, gcd(8,3)=1 ✓

8m+12n = 64+36 = 100

The answer is 100.