Given that
$f(x)=\left\{\begin{array}{ccc}
\frac{x+2}{x^2+3 x+2}, & \text { if } & x \in R-\{-1,-2\} \\
-1, & \text { if } & x=-2 \\
0, & \text { if } & x=-1
\end{array}\right.$
Now, we have to check the continuity
$\begin{aligned}
& \text { at } x= \\
& \text { at } x=-2,-1 \\
& \begin{aligned}
\text { LHL } & =\lim _{h \rightarrow 0} \frac{(-2-h)+2}{(-2-h)^2+3(-2-h)+2} \\
& =\lim _{h \rightarrow 0} \frac{-h}{4+h^2+4 h-6-3 h+2} \\
& =\lim _{h \rightarrow 0} \frac{-h}{h^2+h}=\lim _{h \rightarrow 0} \frac{-1}{h+1}=-1
\end{aligned}
\end{aligned}$
$\begin{aligned}
\text { RHL } & =\lim _{h \rightarrow 0} \frac{(-2+h)+2}{(-2+h)^2+3(-2+h)+2} \\
& =\lim _{h \rightarrow 0} \frac{h}{4+h^2-4 h-6+3 h+2} \\
& =\lim _{h \rightarrow 0} \frac{h}{h^2-h} \\
& =\lim _{h \rightarrow 0} \frac{1}{h-1}=-1 \\
\Rightarrow \text { LHL } & =\text { RHL }=f(-2)
\end{aligned}$
$\therefore$ It is continuous at $x=-2$
Now, check for $x=-1$
$\begin{aligned}
\text { LHL } & =\lim _{h \rightarrow 0} \frac{(-1-h)+2}{(-1-h)^2+3(-1-h)+2} \\
& =\lim _{h \rightarrow 0} \frac{1-h}{1^2+h^2+2 h-3-3 h+2} \\
& =\lim _{h \rightarrow 0} \frac{1-h}{h^2-h}=\lim _{h \rightarrow 0} \frac{-1}{2 h-1}=1 \\
\text { RHL } & =\lim _{h \rightarrow 0} \frac{(-1+h)+2}{(-1+h)^2+3(-1+h)+2} \\
& =\lim _{h \rightarrow 0} \frac{1+h}{1+h^2-2 h-3+3 h+2} \\
& =\lim _{h \rightarrow 0} \frac{1+h}{h^2+h}=1
\end{aligned}$
$\Rightarrow \mathrm{LHL}=\mathrm{RHL} \neq f(-1)$
$\therefore$ It is not continuous at $x=-1$
The required function is continuous in $R-\{-1\}$.