To show that f(x)=absxf(x)=|x| is continuous at 00, 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.