Is the function defined by `f(x) = |x|`, a continuous function?

Explanation

`f(x) = |x|`


`f(x)={-x,if x<0; x, if x>=0`


Since we need to find continuity of the function

We will check continuity for different values of x

`Rightarrow` When `x=0`

`Rightarrow` When `x<0`

`Rightarrow` When `x>0`



Case 1: When `x=0`

`f(x)` is continuous at `x=0`

if L.H.L.=R.H.L.=`f(0)`

i.e. `underset(xto0^-)(lim)f(x)=underset(xto0^+)(lim)f(x)=f(0)`


L.H.L. at `xto0`

`underset(xto0^-)(lim)f(x)=underset(hto0)(lim)f(0-h)`

`underset(xto0^-)(lim)f(x)=underset(hto0)(lim)f(-h)`

`underset(xto0^-)(lim)f(x)=underset(hto0)(lim)|-h|`

`underset(xto0^-)(lim)f(x)=underset(hto0)(lim)h`

`underset(xto0^-)(lim)f(x)=0`


R.H.L. at `xto0`

`underset(xto0^+)(lim)f(x)=underset(hto0)(lim)f(0+h)`

`underset(xto0^+)(lim)f(x)=underset(hto0)(lim)f(h)`

`underset(xto0^+)(lim)f(x)=underset(hto0)(lim)|h|`

`underset(xto0^+)(lim)f(x)=underset(hto0)(lim)h`

`underset(xto0^+)(lim)f(x)=0`


And `f(0)=0`

Hence, L.H.L. = R.H.L. = `f(0)`

`therefore` `f` is continuous at `x=0`


Case 2: When `x<0`

For `x<0`,

`f(x)=-x`

Since this is a polynomial

It is continuous

`therefore` `f(x)` is continuous for `x<0`


Case 3: When `x>0`

For `x>0`,

`f(x)=x`

Since this is a polynomial

It is continuous

`therefore` `f(x)` is continuous for `x>0`


Hence, `f(x)=|x|` is continuous at all points.

`f` is continuous at `x` `varepsilon` `RR`