Q. 85 - Calculus

Question

Use the delta-epsilon definition of continuity to argue that f is or is not continuous at the indicated point $x=c$.

$f\left(x\right)=\left\{\begin{array}{r}2-x,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }x\text{ rational }\\ x^2,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }x\text{ irrational },\end{array}c=2\right.$

Step-by-Step Solution

Answer

Ans:  $f\left(x\right)=\left\{\begin{array}{l}2-x,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }x\text{ rational }\\ x^2,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }x\text{ irrational },\end{array}c=2\right.$ is not continuous at $c=2$

Step 1. Given information.

given,

           $f\left(x\right)=\left\{\begin{array}{r}2-x,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }x\text{ rational }\\ x^2,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }x\text{ irrational },\end{array}c=2\right.$

Step 2. Finding the limits at point c = 2

LHL $=\underset{x\to 2^{-}}{lim} f\left(x\right) =\underset{x\to 2^{-}}{lim} 2-x =2-2 =0$

RHL $=$$\underset{x\to 2^{+}}{lim} f\left(x\right) =\underset{x\to 2^{+}}{lim} x^2 =2^2 =4$$= \underset{x\to 1^{-}}{lim} f\left(x\right) =\underset{x\to 1^{-}}{lim} 2-x =2-1 =1$

Since LHL$\ne$RHL

Therefore, $f\left(x\right)=\left\{\begin{array}{l}2-x,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }x\text{ rational }\\ x^2,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }x\text{ irrational },\end{array}c=2\right.$ is not continuous.