`lim_(x rarr 2)sum_(n=1)^9x/(n(n+1)x^2+2(2n+1)x+4)`

Answer

To solve the limit problem, we will follow these steps:

**Step 1: Substitute the limit value into the expression.**
We need to evaluate the limit:
\[ \lim_{x \to 2} \sum_{n=1}^{9} \frac{x}{n(n+1)x^2 + 2(2n+1)x + 4} \]
Substituting \(x = 2\):
\[ \sum_{n=1}^{9} \frac{2}{n(n+1)(2^2) + 2(2(2n+1)) + 4} \]
Calculating \(2^2 = 4\):
\[ = \sum_{n=1}^{9} \frac{2}{n(n+1)(4) + 4(2n + 1) + 4} \]
This simplifies to:
\[ = \sum_{n=1}^{9} \frac{2}{4n(n+1) + 8n + 4 + 4} \]
\[ = \sum_{n=1}^{9} \frac{2}{4n(n+1) + 8n + 8} \]

**Step 2: Simplify the denominator.**
The denominator can be simplified:
\[ = \sum_{n=1}^{9} \frac{2}{4(n^2 + 3n + 2)} \]
Factoring the quadratic:
\[ = \sum_{n=1}^{9} \frac{2}{4(n+1)(n+2)} \]
This simplifies to:
\[ = \frac{1}{2} \sum_{n=1}^{9} \frac{1}{(n+1)(n+2)} \]

**Step 3: Use partial fraction decomposition.**
We can express \(\frac{1}{(n+1)(n+2)}\) using partial fractions:
\[ \frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2} \]
Thus, the sum becomes:
\[ = \frac{1}{2} \sum_{n=1}^{9} \left( \frac{1}{n+1} - \frac{1}{n+2} \right) \]

**Step 4: Evaluate the telescoping series.**
This series telescopes:
\[ = \frac{1}{2} \left( \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{10} - \frac{1}{11} \right) \right) \]
Most terms cancel out, leaving:
\[ = \frac{1}{2} \left( \frac{1}{2} - \frac{1}{11} \right) \]

**Step 5: Calculate the final result.**
Finding a common denominator:
\[ = \frac{1}{2} \left( \frac{11 - 2}{22} \right) = \frac{1}{2} \cdot \frac{9}{22} = \frac{9}{44} \]

Thus, the final result is:
\[ \lim_{x \to 2} \sum_{n=1}^{9} \frac{x}{n(n+1)x^2 + 2(2n+1)x + 4} = \frac{9}{44} \]