# Find whether $f(x)$ is continuous or discontinuous at the indicated point $f(x)=\left \{\begin{array}{1 1}\large\frac{x^2}{2}, & if\;0\leq x\leq 1\\2x^2-3x+\large\frac{3}{2}\normalsize,& if\;0< x\leq 2\end{array}\right.$ at $x=1$.

$\begin{array}{1 1} Continuous \\ Discontinuous \end{array}$

Toolbox:
• A function is said to be continuous at a point $'a'$ if the LHL = RHL.
• (i.e) $\lim\limits_{\large x\to a^+}f(x)=\lim\limits_{\large x\to a^-}f(x)$
Step 1:
Given : $f(x)=\left \{\begin{array}{1 1}\large\frac{x^2}{2}, & if\;0\leq x\leq 1\\2x^2-3x+\large\frac{3}{2}\normalsize,& if\;0< x\leq 2\end{array}\right.$ at x=1.
The given function is $\large\frac{x^2}{2}$(between the intervals $0$ and $1$)
The other function is $(2x^2-3x+\large\frac{3}{2})$(between 1 and 2)
Step 2:
The left hand limit :
$\lim\limits_{\large x\to 1^-}f(x)=\lim\limits_{\large h\to 0}f(1-h)$
$\qquad\qquad=\lim\limits_{\large h\to 0}\large\frac{(1-h)^2}{2}$
$\qquad\qquad=\lim\limits_{\large h\to 0}\large\frac{1-2h+h^2}{2}$
$\qquad\qquad=\lim\limits_{\large h\to 0}\large\frac{1}{2}$$-h+\large\frac{h^2}{2}$
On applying limits we get,
$\qquad\qquad=1$
Step 3:
Now $(2x^2-3x+\large\frac{3}{2})$ is on the RHS of $x=1$
$\lim\limits_{\large x\to 1^+}f(x)=\lim\limits_{h\to 0}f(1+h)$
$\qquad\qquad=\lim\limits_{\large h\to 0}2(1+h)-3(1+h)+\large\frac{3}{2}$
$\qquad\qquad=\lim\limits_{\large h\to 0}2+2h^2+4h-3+3h+\large\frac{3}{2}$
On applying the limits we get,
$\qquad\qquad=\large\frac{1}{2}$
Hence $\lim\limits_{\large x\to 1^-}f(x)=\lim\limits_{\large x\to 1^+}=\large\frac{1}{2}$
This proves the continuity of the function at $x=1$