Question

If $f\left(x\right)=\frac{x^2}{2},0\le x<1$ and$f\left(x\right)=\frac{2x^2-2x+3}{2};1\le x\le 2$, then $\underset{x\to 1}{lim}f\left(x\right)=$

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. does not exist
• D. $-\frac{1}{2}$

Answer 1

The correct option is C does not exist

For $0\le x<1$, $f\left(x\right)=\frac{x^2}{2}$

Left hand limit of f(x) at 1 $=\underset{x\to 1^{-}}{lim}f\left(x\right)=\frac{1}{2}$

For $1\le x\le 2$, $f\left(x\right)=\frac{2x^2-2x+3}{2}$

Right hand limit of f(x) at 1 $=\underset{x\to 1^{+}}{lim}f\left(x\right)=\frac{3}{2}$

Since the left hand limit and right hand limit at 1 do not coincide, the limit of function f(x) at 1 does not exist.