# The value of

Question:

$f(x)=\left\{\begin{array}{l}\frac{x^{2}}{2}, \text { if } 0 \leq x \leq 1 \\ 2 x^{2}-3 x+\frac{3}{2}, \text { if } 1 Solution: Checking the right hand and left hand limits for the given function, we have$\lim _{x \rightarrow 1^{-}} f(x)=\frac{x^{2}}{2}=\lim _{h \rightarrow 0} \frac{(1-h)^{2}}{2}=\frac{1}{2}\lim _{x \rightarrow 1} f(x)=\frac{x^{2}}{2}=\frac{(1)^{2}}{2}=\frac{1}{2}\lim _{x \rightarrow 1^{-}} f(x)=2 x^{2}-3 x+\frac{3}{2}=2(1)^{2}-3(1)+\frac{3}{2}=2-3+\frac{3}{2}=\frac{1}{2}$Now, as$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x)=\frac{1}{2}\lim _{x \rightarrow 1^{-}} f(x)=\frac{x^{2}}{2}=\lim _{h \rightarrow 0} \frac{(1-h)^{2}}{2}=\frac{1}{2}\lim _{x \rightarrow 1} f(x)=\frac{x^{2}}{2}=\frac{(1)^{2}}{2}=\frac{1}{2}\lim _{x \rightarrow 1^{-}} f(x)=2 x^{2}-3 x+\frac{3}{2}=2(1)^{2}-3(1)+\frac{3}{2}=2-3+\frac{3}{2}=\frac{1}{2}$Now, as$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x)=\frac{1}{2}\$

Thus, the given function f(x) is continuous at x = 1.