Question

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 $\text{gcd}(m, n) = 1$, then $8m + 12n$ is equal to _____

Answer 1

Consider the limit:
$\lim_{x \to 1} \frac{(5x + 1)^{1/3} - (x + 5)^{1/3}}{(2x + 3)^{1/2} - (x + 4)^{1/2}}.$
Using the first-order Taylor expansion for small differences around $x = 1$:
Expand $(5x + 1)^{1/3}$ and $(x + 5)^{1/3}$ around $x = 1$:
$(5x + 1)^{1/3} \approx (5 \times 1 + 1)^{1/3} + \frac{1}{3} \times (5)^{1/3} \times (x - 1),$ $(x + 5)^{1/3} \approx (1 + 5)^{1/3} + \frac{1}{3} \times (1)^{1/3} \times (x - 1).$
The difference becomes: $(5x + 1)^{1/3} - (x + 5)^{1/3} \approx \frac{1}{3} \left(5^{1/3} - 1\right) (x - 1).$
Similarly, expand $(2x + 3)^{1/2}$ and $(x + 4)^{1/2}$:
$(2x + 3)^{1/2} \approx (2 \times 1 + 3)^{1/2} + \frac{1}{2} \times (2)^{1/2} \times (x - 1),$
$(x + 4)^{1/2} \approx (1 + 4)^{1/2} + \frac{1}{2} \times (1)^{1/2} \times (x - 1).$
The difference becomes: $(2x + 3)^{1/2} - (x + 4)^{1/2} \approx \frac{1}{2} \left(2^{1/2} - 1\right) (x - 1).$
Substitute the expansions into the limit:
$\lim_{x \to 1} \frac{\frac{1}{3} \left(5^{1/3} - 1\right) (x - 1)}{\frac{1}{2} \left(2^{1/2} - 1\right) (x - 1)} = \frac{\frac{1}{3} \left(5^{1/3} - 1\right)}{\frac{1}{2} \left(2^{1/2} - 1\right)} = \frac{8 \sqrt{5}}{3 \times 6^{2/3}}.$
Comparing with the given expression:
$\frac{m \sqrt{5}}{n(2n)^{2/3}} \implies m = 8, \quad n = 3.$
Calculating $8m + 12n$:
$8m + 12n = 8 \times 8 + 12 \times 3 = 64 + 36 = 100.$