`lim_(xrarroo)((1^(4)+2^(4)+3^(4)+….+x^(4)))/(x^(2)(1^(2)+2^(2)+……+x^(2)))` equals

Options

A. `(1)/(5)`
B. `(2)/(5)`
C. `(3)/(5)`
D. None of these

Answer

To solve the limit

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

we will first use the formulas for the sums of powers of natural numbers.

### Step 1: Identify the sums
The sum of the first \( n \) natural numbers raised to the fourth power is given by:

\[ \sum_{k=1}^n k^4 = \frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30}. \]

The sum of the first \( n \) natural numbers raised to the second power is given by:

\[ \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}. \]

### Step 2: Substitute the sums into the limit
Substituting these formulas into our limit gives us:

\[ \lim_{x \to \infty} \frac{\frac{x(x+1)(2x+1)(3x^2 + 3x - 1)}{30}}{x^2 \cdot \frac{x(x+1)(2x+1)}{6}}. \]

### Step 3: Simplify the expression
Now, we simplify the expression:

\[ = \lim_{x \to \infty} \frac{\frac{x(x+1)(2x+1)(3x^2 + 3x - 1)}{30}}{\frac{x^3(x+1)(2x+1)}{6}}. \]

This simplifies to:

\[ = \lim_{x \to \infty} \frac{6(3x^2 + 3x - 1)}{30x} = \lim_{x \to \infty} \frac{3(3x^2 + 3x - 1)}{5x}. \]

### Step 4: Further simplification
Now, we can divide the numerator and denominator by \( x \):

\[ = \lim_{x \to \infty} \frac{3(3x + 3 - \frac{1}{x})}{5}. \]

### Step 5: Apply the limit
As \( x \to \infty \), \( \frac{1}{x} \to 0 \):

\[ = \frac{3(3 \cdot \infty + 3 - 0)}{5} = \frac{3 \cdot 3}{5} = \frac{9}{5}. \]

### Final Answer
Thus, the limit evaluates to:

\[ \frac{9}{5}. \]