1. We need to evaluate

$$\lim_{x\to 2}\sum_{n=1}^9 \frac{x}{n(n+1)x^2+2(2n+1)x+4}.$$

2. Since each term is a rational function and the denominator is nonzero at $x=2$, we can substitute $x=2$ directly.

So for the $n$th term,

$$\frac{x}{n(n+1)x^2+2(2n+1)x+4}\Bigg|_{x=2} = \frac{2}{n(n+1)(4)+2(2n+1)(2)+4}.$$

3. Simplify the denominator:

$$4n(n+1)+4(2n+1)+4 =4n^2+4n+8n+4+4 =4n^2+12n+8.$$

Thus,

$$\frac{2}{4n^2+12n+8} =\frac{2}{4(n^2+3n+2)} =\frac{2}{4(n+1)(n+2)} =\frac{1}{2(n+1)(n+2)}.$$

Hence the limit becomes

$$\sum_{n=1}^9 \frac{1}{2(n+1)(n+2)} =\frac12\sum_{n=1}^9 \frac{1}{(n+1)(n+2)}.$$

4. Use partial fractions:

$$\frac{1}{(n+1)(n+2)}=\frac{1}{n+1}-\frac{1}{n+2}.$$

Therefore,

$$\frac12\sum_{n=1}^9\left(\frac{1}{n+1}-\frac{1}{n+2}\right).$$

5. This is a telescoping sum:

$$\frac12\left[\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\cdots+\left(\frac1{10}-\frac1{11}\right)\right].$$

All middle terms cancel, leaving

$$\frac12\left(\frac12-\frac1{11}\right).$$

6. Compute:

$$\frac12\left(\frac{11-2}{22}\right) =\frac12\cdot\frac9{22} =\frac9{44}.$$

7. So the correct option is

$$\boxed{\frac{9}{44}}.$$

This corresponds to Option A.