Question

How to prove that `f(x)=|x|` is continue at `0` ?

Answer 1

Please see below.

Explanation:

We know that,

`color(blue)((1)If , lim_(xtoa) f(x)=f(a)=>f` is continuos at `color(blue)(x=a`

OR

`color(red)(lim_(xtoa^-) f(x)=lim_(xtoa^+) f(x)=f(a)=>f` is continuos at `color(red)(x=a`

`color(violet)((2)|x|=x, x > 0`

`color(white)(.......)color(violet)(=-x , x < 0`

We have,

`color(red)(f(0)=|0|=0...to(I)`

`lim_(xto0^+)f(x)=lim_(xto0^+) |x|`

`color(white)(................)=lim_(xto0^+) x ...to(x >0 )`

`color(red)(lim_(xto0^+)f(x)=0...to (II)`

`lim_(xto0^-)f(x)=lim_(xto0^-) |x|`

`color(white)(................)=lim_(xto0^-) (-x) ...to(x < 0 )`

`color(red)(lim_(xto0^-)f(x)=0...to (III)`

From , `color(red)((I),(II), and (III)`

`color(green)(lim_(xto0^+)f(x)=lim_(xto0^-)f(x)=f(0)=0`

Hence, `f(x)=|x|` is continuous at `x=0`