The value of
`lim_(xto oo)(2x^(1//2)+3x^(1//3)+4x^(1//4)+....nx^(1//n))/((2x-3)^(1//2)+(2x-3)^(1//3)+....+(2x-3)^(1//n))`is

Options

A. `sqrt(2)`
B. `2`
C. `1/(sqrt(3)`
D. `0`

Answer

To solve the limit

\[ \lim_{x \to \infty} \frac{2x^{1/2} + 3x^{1/3} + 4x^{1/4} + \ldots + nx^{1/n}}{(2x - 3)^{1/2} + (2x - 3)^{1/3} + \ldots + (2x - 3)^{1/n}}, \]

we will follow these steps:

### Step 1: Identify the highest power of \( x \) in the numerator and denominator.

In the numerator, the term with the highest power is \( 2x^{1/2} \). In the denominator, the highest power is \( (2x)^{1/2} \) since \( (2x - 3)^{1/n} \) behaves like \( (2x)^{1/n} \) as \( x \to \infty \).

### Step 2: Divide both the numerator and denominator by \( x^{1/2} \).

This gives us:

\[ \lim_{x \to \infty} \frac{2 + 3x^{-1/6} + 4x^{-1/4} + \ldots + nx^{(1/n) - (1/2)}}{(2 - \frac{3}{x})^{1/2} + (2 - \frac{3}{x})^{1/3} + \ldots + (2 - \frac{3}{x})^{1/n}}. \]

### Step 3: Simplify the numerator.

As \( x \to \infty \), all terms involving \( x^{-k} \) (where \( k > 0 \)) in the numerator tend to 0. Thus, the numerator approaches:

\[ 2. \]

### Step 4: Simplify the denominator.

For the denominator, we can approximate \( (2 - \frac{3}{x})^{1/k} \) as \( 2^{1/k} \) since \( \frac{3}{x} \to 0 \) as \( x \to \infty \). Therefore, the denominator becomes:

\[ 2^{1/2} + 2^{1/3} + \ldots + 2^{1/n}. \]

### Step 5: Evaluate the limit.

Now, we can rewrite the limit as:

\[ \lim_{x \to \infty} \frac{2}{2^{1/2} + 2^{1/3} + \ldots + 2^{1/n}}. \]

### Step 6: Find the value of the denominator.

The denominator can be approximated as \( n \cdot 2^{1/n} \) for large \( n \) since each term \( 2^{1/k} \) approaches 1 as \( k \) increases.

### Step 7: Final limit calculation.

Thus, the limit simplifies to:

\[ \frac{2}{2^{1/2} + 2^{1/3} + \ldots + 2^{1/n}} \approx \frac{2}{\text{(sum of terms that approach 1)}}. \]

As \( n \to \infty \), the sum approaches \( n \cdot 1 \), leading us to:

\[ \lim_{x \to \infty} \frac{2}{\sqrt{2}} = \sqrt{2}. \]

### Final Answer:

The value of the limit is

\[ \sqrt{2}. \]