Question:
$\lim _{x \rightarrow 2}\left(\sum_{n=1}^{9} \frac{x}{n(n+1) x^{2}+2(2 n+1) x+4}\right)$ is equal to :
Correct Option: 1
Solution:
$S=\lim _{x \rightarrow 2} \sum_{n=1}^{9} \frac{x}{n(n+1) x^{2}+2(2 n+1) x+4}$
$\mathrm{S}=\sum_{\mathrm{n}=1}^{9} \frac{2}{4\left(\mathrm{n}^{2}+3 \mathrm{n}+2\right)}=\frac{1}{2} \sum_{\mathrm{n}=1}^{9}\left(\frac{1}{\mathrm{n}+1}-\frac{1}{\mathrm{n}+2}\right)$
$\mathrm{S}=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{11}\right)=\frac{9}{44}$
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