text: ## Calculus (3rd Edition)
  
  $$\lim _{x \rightarrow-\frac{1}{2}} \frac{2 x+1}{2 x^{2}+3 x+1}=2$$
  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}