Key Concepts and Formulas

• Limit of a Sum: If $\mathop {\lim }\limits_{x \to a} f(x)$ and $\mathop {\lim }\limits_{x \to a} g(x)$ exist, then $\mathop {\lim }\limits_{x \to a} (f(x) + g(x)) = \mathop {\lim }\limits_{x \to a} f(x) + \mathop {\lim }\limits_{x \to a} g(x)$. In this case, we can interchange the limit and the summation if the limit of the summand exists.
• Telescoping Series: A series where most of the terms cancel out, leaving only the first and last terms. A common form is $\sum_{n=1}^N (a_n - a_{n+1})$ or $\sum_{n=1}^N (a_{n+1} - a_n)$.
• Partial Fraction Decomposition: A technique used to express a rational function as a sum of simpler rational functions. For example, $\frac{1}{(n+1)(n+2)} = \frac{A}{n+1} + \frac{B}{n+2}$.

Step-by-Step Solution

Step 1: Evaluate the limit of the summand as $x \to 2$. The problem asks for the limit of a sum. We can first evaluate the limit of the general term inside the summation as $x$ approaches 2. This is because the summand is a rational function of $x$, and as long as the denominator is non-zero at $x=2$, we can directly substitute $x=2$.

Let the general term be $T_n(x) = \frac{x}{n(n + 1)x^2 + 2(2n + 1)x + 4}$. We need to find $\mathop {\lim }\limits_{x \to 2} T_n(x)$. Substituting $x=2$ into the denominator: $n(n + 1)(2)^2 + 2(2n + 1)(2) + 4$ $= 4n(n + 1) + 4(2n + 1) + 4$ $= 4(n^2 + n) + 8n + 4 + 4$ $= 4n^2 + 4n + 8n + 8$ $= 4n^2 + 12n + 8$ $= 4(n^2 + 3n + 2)$ $= 4(n+1)(n+2)$. Since $n$ ranges from 1 to 9, $(n+1)(n+2)$ will never be zero. Thus, the denominator is non-zero at $x=2$.

So, $\mathop {\lim }\limits_{x \to 2} T_n(x) = \frac{2}{4(n+1)(n+2)} = \frac{1}{2(n+1)(n+2)}$.

Step 2: Interchange the limit and the summation. Since the limit of each term in the sum exists, we can move the limit inside the summation: $S = \mathop {\lim }\limits_{x \to 2} \sum\limits_{n = 1}^9 {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}} = \sum\limits_{n = 1}^9 {\mathop {\lim }\limits_{x \to 2} {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}}}$ From Step 1, we know that $\mathop {\lim }\limits_{x \to 2} {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}} = \frac{1}{2(n+1)(n+2)}$. Therefore, the problem reduces to evaluating the sum: $S = \sum\limits_{n = 1}^9 {\frac{1}{2(n+1)(n+2)}}$

Step 3: Simplify the general term of the sum using partial fraction decomposition. The general term of the sum is $\frac{1}{2(n+1)(n+2)}$. We can factor out the constant $\frac{1}{2}$ and focus on decomposing $\frac{1}{(n+1)(n+2)}$. We want to find constants A and B such that: $\frac{1}{(n+1)(n+2)} = \frac{A}{n+1} + \frac{B}{n+2}$ Multiplying both sides by $(n+1)(n+2)$, we get: $1 = A(n+2) + B(n+1)$ To find A, let $n = -1$: $1 = A(-1+2) + B(-1+1) \implies 1 = A(1) + B(0) \implies A = 1$. To find B, let $n = -2$: $1 = A(-2+2) + B(-2+1) \implies 1 = A(0) + B(-1) \implies 1 = -B \implies B = -1$. So, $\frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2}$ Therefore, the general term of the sum can be written as: $\frac{1}{2(n+1)(n+2)} = \frac{1}{2}\left(\frac{1}{n+1} - \frac{1}{n+2}\right)$

Step 4: Evaluate the sum using the telescoping property. Now, we substitute this back into the summation: $S = \sum\limits_{n = 1}^9 {\frac{1}{2}\left(\frac{1}{n+1} - \frac{1}{n+2}\right)}$ We can pull the constant $\frac{1}{2}$ out of the summation: $S = \frac{1}{2} \sum\limits_{n = 1}^9 {\left(\frac{1}{n+1} - \frac{1}{n+2}\right)}$ This is a telescoping series. Let's write out the first few terms and the last term to see the cancellation: For $n=1$: $\frac{1}{2} - \frac{1}{3}$ For $n=2$: $\frac{1}{3} - \frac{1}{4}$ For $n=3$: $\frac{1}{4} - \frac{1}{5}$ ... For $n=9$: $\frac{1}{10} - \frac{1}{11}$

Summing these terms: $\sum\limits_{n = 1}^9 {\left(\frac{1}{n+1} - \frac{1}{n+2}\right)} = \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \dots + \left(\frac{1}{10} - \frac{1}{11}\right)$ Notice that the $-\frac{1}{3}$ cancels with $+\frac{1}{3}$, $-\frac{1}{4}$ cancels with $+\frac{1}{4}$, and so on. This pattern continues until the second to last term. The only terms that remain are the first part of the first term and the second part of the last term. $\sum\limits_{n = 1}^9 {\left(\frac{1}{n+1} - \frac{1}{n+2}\right)} = \frac{1}{2} - \frac{1}{11}$

Step 5: Calculate the final value of the sum. Now, we substitute this result back into the expression for S: $S = \frac{1}{2} \left(\frac{1}{2} - \frac{1}{11}\right)$ To subtract the fractions inside the parenthesis, find a common denominator, which is $2 \times 11 = 22$: $\frac{1}{2} - \frac{1}{11} = \frac{1 \times 11}{2 \times 11} - \frac{1 \times 2}{11 \times 2} = \frac{11}{22} - \frac{2}{22} = \frac{11 - 2}{22} = \frac{9}{22}$ Finally, multiply by $\frac{1}{2}$: $S = \frac{1}{2} \times \frac{9}{22} = \frac{9}{44}$

Common Mistakes & Tips

• Incorrectly Interchanging Limit and Summation: Always ensure that the limit of the summand exists before interchanging the limit and the summation. If the limit of the summand does not exist, this step is invalid.
• Errors in Partial Fraction Decomposition: Double-check the values of the constants A and B. A common mistake is to miscalculate these values, which will lead to an incorrect telescoping sum.
• Mistakes in Telescoping Sum Calculation: Carefully identify the terms that cancel out. Ensure that the first term's positive part and the last term's negative part are correctly identified.

Summary

The problem requires evaluating a limit of a sum. We first evaluated the limit of the general term of the sum as $x$ approaches 2, which simplified the expression to a sum involving $n$. We then used partial fraction decomposition to rewrite the general term of the sum in a form suitable for a telescoping series. By writing out the terms of the series, we observed that most terms cancel out, leaving only the first and last terms. Finally, we computed the value of the remaining terms to arrive at the final answer.

The final answer is \boxed{\frac{9}{44}}. This corresponds to option (A).