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

Answer 2

Step 1: Write down the given limit.
We are given: \[ \lim_{x \to 1} \frac{(5x + 1)^{1/3} - (x + 5)^{1/3}}{(2x + 3)^{1/2} - (x + 4)^{1/2}} = \frac{m\sqrt{5}}{n(2n)^{2/3}}. \] We need to find \( 8m + 12n \), given that \( \text{gcd}(m, n) = 1 \).

Step 2: Recognize that it is an indeterminate form.
When \( x = 1 \):
Numerator = \( (6)^{1/3} - (6)^{1/3} = 0 \)
Denominator = \( (5)^{1/2} - (5)^{1/2} = 0 \)
Hence, the expression is of the indeterminate form \( \frac{0}{0} \). We can apply **L’Hôpital’s Rule**.

Step 3: Apply L’Hôpital’s Rule.
Differentiate numerator and denominator with respect to \( x \):

Numerator derivative: \[ \frac{d}{dx}\left[(5x + 1)^{1/3} - (x + 5)^{1/3}\right] = \frac{1}{3}(5x + 1)^{-2/3} \cdot 5 - \frac{1}{3}(x + 5)^{-2/3}. \] Simplify: \[ \text{Numerator derivative} = \frac{5(5x + 1)^{-2/3} - (x + 5)^{-2/3}}{3}. \] Denominator derivative: \[ \frac{d}{dx}\left[(2x + 3)^{1/2} - (x + 4)^{1/2}\right] = \frac{1}{2}(2x + 3)^{-1/2} \cdot 2 - \frac{1}{2}(x + 4)^{-1/2}. \] Simplify: \[ \text{Denominator derivative} = (2x + 3)^{-1/2} - \frac{1}{2}(x + 4)^{-1/2}. \]

Step 4: Evaluate at \( x = 1 \).
Substitute \( x = 1 \):
\[ (5x + 1) = 6, \quad (x + 5) = 6, \quad (2x + 3) = 5, \quad (x + 4) = 5. \] So: \[ \text{Numerator derivative} = \frac{5(6)^{-2/3} - (6)^{-2/3}}{3} = \frac{4(6)^{-2/3}}{3}. \] \[ \text{Denominator derivative} = (5)^{-1/2} - \frac{1}{2}(5)^{-1/2} = \frac{1}{2}(5)^{-1/2}. \]

Step 5: Simplify the limit.
\[ L = \frac{\text{Numerator derivative}}{\text{Denominator derivative}} = \frac{\frac{4(6)^{-2/3}}{3}}{\frac{1}{2}(5)^{-1/2}} = \frac{4(6)^{-2/3}}{3} \cdot \frac{2(5)^{1/2}}{1} = \frac{8\sqrt{5}}{3(6)^{2/3}}. \] Simplify \( (6)^{2/3} \):
\[ (6)^{2/3} = (2 \cdot 3)^{2/3} = (2^{2/3})(3^{2/3}). \] So: \[ L = \frac{8\sqrt{5}}{3 \cdot 2^{2/3} \cdot 3^{2/3}} = \frac{8\sqrt{5}}{3^{5/3} \cdot 2^{2/3}}. \]

Step 6: Match with the given form.
Given: \[ L = \frac{m\sqrt{5}}{n(2n)^{2/3}}. \] Comparing: \[ \frac{m\sqrt{5}}{n(2n)^{2/3}} = \frac{8\sqrt{5}}{3^{5/3} \cdot 2^{2/3}}. \] We can rewrite denominator on RHS as: \[ 3^{5/3} \cdot 2^{2/3} = (3 \cdot (2 \cdot 3)^{2/3}) = 3 \cdot (6)^{2/3}. \] Thus, we can identify \( n = 3 \) because \( 2n = 6 \).
Now, numerator comparison gives \( m = 8 \).

Step 7: Compute \( 8m + 12n \).
\[ 8m + 12n = 8(8) + 12(3) = 64 + 36 = 100. \]

Final Answer:
\[ \boxed{100} \]