Question

Discuss the continuity of the f(x) at the indicated points:
(i) f(x) = | x | + | x − 1 | at x = 0, 1.
(ii) f(x) = | x − 1 | + | x + 1 | at x = −1, 1.

Answer 1

(i) Given: $f\left(x\right)=\left|x\right|+\left|x-1\right|$

We have
(LHL at x = 0) = $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0-h\right)$$=\underset{h\to 0}{\lim }\left[\left|0-h\right|+\left|0-h-1\right|\right]=1$

(RHL at x = 0) = $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0+h\right)$$=\underset{h\to 0}{\lim }\left[\left|0+h\right|+\left|0+h-1\right|\right]=1$

Also, $f\left(0\right)=\left|0\right|+\left|0-1\right|=0+1=1$

Now,

(LHL at x = 1) = $\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1-h\right)=\underset{h\to 0}{\lim }\left(\left|1-h\right|+\left|1-h-1\right|\right)=1+0=1$

(RHL at x =1) = $\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1+h\right)=\underset{h\to 0}{\lim }\left(\left|1+h\right|+\left|1+h-1\right|\right)=1+0=1$

Also, $f\left(1\right)=\left|1\right|+\left|1-1\right|=1+0=1$

∴ ​$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=f\left(0\right)\mathrm{and}\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)=f\left(1\right)$

Hence, $f\left(x\right)$ is continuous at $x=0,1$.

(ii) Given: $f\left(x\right)=\left|x-1\right|+\left|x+1\right|$

We have
(LHL at x = −1) = $\underset{x\to -1^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(-1-h\right)$$=\underset{h\to 0}{\lim }\left[\left|-1-h-1\right|+\left|-1-h+1\right|\right]=2+0=2$

(RHL at x = −1) = $\underset{x\to -1^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(-1+h\right)$$=\underset{h\to 0}{\lim }\left[\left|-1+h-1\right|+\left|-1+h+1\right|\right]=2+0=2$

Also, $f\left(-1\right)=\left|-1-1\right|+\left|-1+1\right|=\left|-2\right|=2$

Now,

(LHL at x = 1) = $\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1-h\right)=\underset{h\to 0}{\lim }\left(\left|1-h-1\right|+\left|1-h+1\right|\right)=0+2=2$

(RHL at x =1) = $\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1+h\right)=\underset{h\to 0}{\lim }\left(\left|1+h-1\right|+\left|1+h+1\right|\right)=0+2=2$

Also, $f\left(1\right)=\left|1+1\right|+\left|1-1\right|=2$

∴ ​$\underset{x\to -1^{-}}{\lim }f\left(x\right)=\underset{x\to -1^{+}}{\lim }f\left(x\right)=f\left(-1\right)\mathrm{and}\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)=f\left(1\right)$

Hence, $f\left(x\right)$ is continuous at $x=-1,1$.