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If $f: R \rightarrow R$ defined as $f(x)=\frac{x^3+2 x^2+x+2}{x^2+x-2}$ (when $x \neq-2$ ) is continuous at $x=-2$, then $f(-2)$ is equal to
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The correct answer is:
$\frac{-5}{3}$
Given, $f(x)=\frac{x^3+2 x^2+x+2}{x^2+x-2}$ is continuous at
$x=-2$
$\begin{aligned} & \therefore \quad \lim _{x \rightarrow-2} f(x)=f(-2) \\ & \Rightarrow \quad f(-2)=\lim _{x \rightarrow-2} \frac{(x+2)\left(x^2+1\right)}{(x+2)(x-1)}=\frac{4+1}{-2-1}=\frac{-5}{3}\end{aligned}$
$x=-2$
$\begin{aligned} & \therefore \quad \lim _{x \rightarrow-2} f(x)=f(-2) \\ & \Rightarrow \quad f(-2)=\lim _{x \rightarrow-2} \frac{(x+2)\left(x^2+1\right)}{(x+2)(x-1)}=\frac{4+1}{-2-1}=\frac{-5}{3}\end{aligned}$
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