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

Correct Answer: 100

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 \(8m + 12n\): 
\[ 8m + 12n = 8 \times 8 + 12 \times 3 = 64 + 36 = 100. \]