Mathematics - Limits Question with Solution | TestHub

$\underset{x\to 2 }{\lim }\sum _{n=1}^9\left(\frac{x}{n(n+1)x^2+2(2n+1)x+4}\right)$

$\underset{x\to 2}{\lim }\sum _{ n=1}^9\left(\frac{x}{n\left(n+1\right)x^2+2nx+2\left(n+1\right)x+4}\right)$

$\underset{x\to 2}{\lim }\sum _{ n=1}^9\left(\frac{\left[(n+1)x+2\right]-\left[nx+2\right]}{\left[(n+1)x+2\right]\left[nx+2\right]}\right)$

$\underset{x\to 2}{\lim }\sum _{ n=1}^9\left(\frac{1}{(nx+2)}-\frac{1}{(n+1)x+2}\right)$

$=\frac{1}{x+2}-\frac{1}{2x+2}$

$\frac{1}{2x+2}-\frac{1}{3x+2}$

$$\begin{gathered} \begin{array}{c}.\\ .\\ .\\ .\end{array} \\ \frac{+\frac{1}{9x+2}-\frac{1}{10x+2}}{\frac{1}{x+2}-\frac{1}{10x+2}} \end{gathered}$$

$\Rightarrow \underset{x\to 2}{\lim }\left(\frac{1}{x+2}-\frac{1}{10x+2}\right)=\frac{9}{44}$

$\Rightarrow \underset{x\to 2}{\lim }\frac{10x+2-x-2}{\left(x+2\right)\left(10x+2\right)}$

$\underset{x\to 2}{\lim }\frac{9x}{\left(x+2\right)\left(10x+2\right)}=\frac{18}{4\times 22}=\frac{9}{44}$