If $f:R ightarrow R$ is defined by $f(x) = \left\{ \begin{array}{r} \frac{x + 2}{x^{2} + 3x + 2},\ \ \ if\ x \in R - \{ - 1,\ - 2\} \ - 1,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ if\ x = - 2 \ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ if\ \ x = - 1 \end{array} ight.\$, then $f$ is continuous on the set

Question

If $f:R ightarrow R$ is defined by $f(x) = \left\{ \begin{array}{r} \frac{x + 2}{x^{2} + 3x + 2},\ \ \ if\ x \in R - \{ - 1,\ - 2\} \ - 1,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ if\ x = - 2 \ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ if\ \ x = - 1 \end{array} ight.\$, then $f$ is continuous on the set

Dr. Abubacker · Accepted Answer

(c)
Since, $f(x)$ is continuous for every value of $R$ except {-1, -2}. Now, we have to check that points
At $x = - 2$
LHL$= \lim_{h ightarrow 0}\frac{( - 2 - h) + 2}{( - 2 - h)^{2} + 3( - 2 - h) + 2}$
$= \lim_{h ightarrow 0}\frac{- h}{h^{2} + h} = - 1$
RHL$= \lim_{h ightarrow 0}\frac{( - 2 + h) + 2}{( - 2 + h)^{2} + 3( - 2 + h) + 2}$
$= \lim_{h ightarrow 0}\frac{h}{h^{2} - h} = - 1$
$\Rightarrow \$LHL=RHL=$f( - 2)$
$	herefore$ It is continuous at $x = - 2$
Now, check for $x = - 1$
LHL$= \lim_{h ightarrow 0}\frac{( - 1 - h) + 2}{( - 1 - h)^{2} + 3( - 1 - h) + 2}$
$= \lim_{h ightarrow 0}\frac{1 - h}{h^{2} - h} = \infty$
RHL$= \lim_{h ightarrow 0}\frac{( - 1 + h) + 2}{( - 1 + h)^{2} + 3( - 1 + h) + 2}$
$= \lim_{h ightarrow 0}\frac{1 + h}{h^{2} + h} = \infty$
$\Rightarrow$ LHL=RHL$
eq f( - 1)$
$	herefore$ It is not continuous at $x = - 1$
The required function is continuous in $R - \{ - 1\}$

Explanation

Reviewed by Dr. Abubacker

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