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

1 Answer
Apr 22, 2018

Answer:

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#