Question

Is the function f defined by $f\left(x\right)=\{\begin{array}{c}x,ifx\le 1\\ 5,ifx>1\end{array}$ continuous at x = 0? At x=1? At x =2?

Answer 1

Here, $f\left(x\right)=\{\begin{array}{c}x,ifx\le 1\\ 5,ifx>1\end{array}$

At x=0, LHL=$\underset{x\to 0^{-}}{lim}$ f(x) = $\underset{x\to 0^{-}}{lim}$ (x)

Putting x=0-h as $x\to 0^{-}$ when $h\to 0$ = $\underset{h\to 0}{lim}$ (0=h)=0-0=0

RHL = $\underset{x\to 0^{+}}{lim}$ f(x) = $\underset{x\to 0^{+}}{lim}$ (x)

Putting x=0 + h as $x\to 0^{+}$ when $h\to 0$ = $\underset{h\to 0}{lim}$ (0+h)=0+0=0

Also, f(0)=0 $[\therefore f(x)=x]$

LHL=RHL=f(0)

Thus, f(x) is continuous at x=0.

At x=1, LHL = $\underset{x\to 1^{-}}{lim}$ f(x)=$\underset{x\to 1^{-}}{lim}$(x)

Putting x=1-h as $x\to 1^{-}$ when $h\to 0$ = $\underset{h\to 0}{lim}$ (1-h)=1-1=1

RHL = $\underset{x\to 1^{+}}{lim}$ f(x) =5 $\therefore LHL\ne RHL$

Thus, f(x) is discontinuous at x=1.

At x=2 $\underset{x\to 2^{-}}{lim}$ f(x)=$\underset{x\to 2^{-}}{lim}$ f (x) = 5

Also, f(2) = 5 $\therefore$ LHL = RHL = f(2). Thus, f(x) is continuous at x=2.

Direction (Q. Nos. 6 to 12) Find all point of discontinuity of $f_1$ where f is defined in the question.