Answer
$$\lim _{x \rightarrow-\frac{1}{2}} \frac{2 x+1}{2 x^{2}+3 x+1}=2$$
Work Step by Step
Given $$\lim _{x \rightarrow-\frac{1}{2}} \frac{2 x+1}{2 x^{2}+3 x+1}$$
let $$ f(x) = \frac{2 x+1}{2 x^{2}+3 x+1}$$
Since, we have
$$ f(-\frac{1}{2})= \frac{-1+1}{\frac{1}{2}-\frac{3}{2}+1}=\frac{0}{0}$$
So, transform algebraically and cancel
\begin{aligned}
L&=\lim _{x \rightarrow-\frac{1}{2}} \frac{2 x+1}{2 x^{2}+3 x+1}\\
&=\lim _{x \rightarrow -\frac{1}{2}} \frac{(2x+1)}{(2x+1)(x+1)}\\
&=\lim _{x \rightarrow -\frac{1}{2}} \frac{1}{(x+1)}\\
&= \frac{1}{(-\frac{1}{2}+1)}\\
&= \frac{1}{(\frac{1}{2})}\\
&=2
\end{aligned}
