If `lim_(xrarroo) (8x^3+mx^2)^(1//3)-nx` exists and is equal to 1 , then the vlaue of `(m)/(n)` is

Options

A. `(1)/(6)`
B. 6
C. 3
D. `(1)/(3)`

Explanation

To solve the problem, we need to evaluate the limit:

\[ \lim_{x \to \infty} \left( (8x^3 + mx^2)^{\frac{1}{3}} - nx \right) \]

and determine the values of \( m \) and \( n \) such that this limit exists and equals 1.

### Step 1: Simplify the expression inside the limit
We start with the expression inside the limit:

\[ (8x^3 + mx^2)^{\frac{1}{3}} \]

As \( x \) approaches infinity, the term \( 8x^3 \) will dominate \( mx^2 \). Therefore, we can factor out \( x^3 \):

\[ (8x^3(1 + \frac{m}{8} \cdot \frac{1}{x})^{\frac{1}{3}}) = (8^{\frac{1}{3}} x) \left(1 + \frac{m}{8} \cdot \frac{1}{x}\right)^{\frac{1}{3}} \]

### Step 2: Evaluate the limit of the cube root
Using the binomial expansion for small values of \( \frac{m}{8} \cdot \frac{1}{x} \) as \( x \to \infty \):

\[ \left(1 + \frac{m}{8} \cdot \frac{1}{x}\right)^{\frac{1}{3}} \approx 1 + \frac{1}{3} \cdot \frac{m}{8} \cdot \frac{1}{x} \]

Thus, we can rewrite our limit as:

\[ \lim_{x \to \infty} \left( 2x \left(1 + \frac{1}{3} \cdot \frac{m}{8} \cdot \frac{1}{x}\right) - nx \right) \]

### Step 3: Combine the terms
This simplifies to:

\[ \lim_{x \to \infty} \left( 2x + \frac{m}{12} - nx \right) \]

### Step 4: Set the limit equal to 1
For the limit to exist and equal 1, the coefficient of \( x \) must equal zero:

\[ 2 - n = 0 \implies n = 2 \]

### Step 5: Determine the value of \( m \)
Now we substitute \( n = 2 \) back into the constant term:

\[ \frac{m}{12} = 1 \implies m = 12 \]

### Step 6: Calculate \( \frac{m}{n} \)
Now we can find the value of \( \frac{m}{n} \):

\[ \frac{m}{n} = \frac{12}{2} = 6 \]

### Final Answer
Thus, the value of \( \frac{m}{n} \) is:

\[ \boxed{6} \]