To show that #f(x)=absx# is continuous at #0#, show that #lim_(xrarr0) absx = abs0 = 0#.

Use #epsilon-delta# if required, or use the piecewise definition of absolute value.

#f(x) = absx = {(x,"if",x >= 0),(-x,"if",x < 0):}#

So, #lim_(xrarr0^+) absx = lim_(xrarr0^+)x = 0#

and #lim_(xrarr0^-) absx = lim_(xrarr0^-)(-x) = 0#.

Therefore,

#lim_(xrarr0) absx =0# which is, of course equal to #f(0)#.

To show that #f(x) = absx# is not differentiable, show that

#f'(0) = lim_(hrarr0) (f(0+h)-f(0))/h# does not exists.

Observe that

#lim_(hrarr0) (abs(0+h)-abs(0))/h = lim_(hrarr0)(absh)/h#

But #absh/h = {(1,"if",h > 0),(-1,"if",h < 0):}#,

so the limit from the right is #1#, while the limit from the left is #-1#.

So the two sided limit does not exist.

That is, the derivative does not exist at #x = 0#.