# If the function defined by $f(x) = \left\{ \begin{array}{l l}2x-1, & \quad if { x < 2 } \\ a, & \quad if { x = 2 } \\ x+1, & \quad if { x > 2 } \end{array} \right.$ is continuous at x = 2, find the value of a. Also discuss the continuity of f(x) at x = 3.

Toolbox:
• If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
$f(x) = \left\{ \begin{array}{l l}2x-1, & \quad if { x < 2 } \\ a, & \quad if { x = 2 } \\ x+1, & \quad if { x > 2 } \end{array} \right.$
For $f(x)$ to be continuous at $x=2$,we must have,
$\lim\limits_{x\to 2^-}f(x)=\lim\limits_{x\to 2^+}f(x)=f(2)$
$\Rightarrow \lim\limits_{x\to 2^-}f(x)=\lim\limits_{x\to 2^+}f(x)=a$
$\therefore$ LHL :$\lim\limits_{x\to 2^-}f(x)=\lim\limits_{x\to 2}2x-1$
$\Rightarrow 3$
Step 2:
RHL :
$\lim\limits_{x\to 2^+}f(x)=\lim\limits_{x\to 2}x+1$
$\qquad\qquad=3$
$\therefore a=3$
Step 3:
Let us next discuss the continuity at $x=3$
LHL :$\lim\limits_{x\to 3^-}f(x)=\lim\limits_{x\to 3}2x-1$
$\Rightarrow 5$
Step 4:
RHL:$\lim\limits_{x\to 3^+}f(x)=\lim\limits_{x\to 3}x+1$
$\Rightarrow 4$
$\therefore$LHL $\neq$ RHL
Hence the function is not continuous at $x=3$