If $f:R\to R$ defined by $f\left(x\right)=\{\begin{array}{c}\frac{x-2}{x^2+3x+2}\text{if}x\in R-\left\{-1,-2\right\}\\ 2\text{if}x=-2\\ 1\text{if}x=-1\end{array}$ then $f$ is continuous on the set

$\underset{x\to -2^{+}}{Lt}f\left(x\right)=\underset{x\to -2^{+}}{Lt}\frac{x-2}{x^2+3x+2}=\underset{x\to -2^{+}}{Lt}\frac{x-2}{(x+1)(x+2)}\to \infty$ 

$\underset{x\to -2^{-}}{Lt}f\left(x\right)=\underset{x\to -2^{-}}{Lt}\frac{(x-2)}{x^2+3x+2}=\underset{x\to -2^{-}}{Lt}\frac{x-2}{(x+1)(x+2)}\to -\infty$

LHL$\ne$RHL

Limit does not exist

similarly at x= - 1 limit does not exist  

$\therefore$  $f$ is continuous an $R–\left\{–1,–2\right\}$