How do you prove that the function f(x) = | x | is continuous at x=0, but not differentiable at x=0?

Question

Charles Anctil · Accepted Answer

See the explanation, below.

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#.

Ezra Adams · Answer

undefined To prove that the function f(x) = |x| is continuous at x=0, we need to show that the limit of f(x) as x approaches 0 exists and is equal to f(0). To do this, we consider the left-hand limit and the right-hand limit separately. For the left-hand limit, as x approaches 0 from the left side (x < 0), f(x) = -x. Taking the limit as x approaches 0 from the left, we have lim(x→0-) -x = 0. For the right-hand limit, as x approaches 0 from the right side (x > 0), f(x) = x. Taking the limit as x approaches 0 from the right, we have lim(x→0+) x = 0. Since both the left-hand limit and the right-hand limit are equal to 0, we can conclude that the limit of f(x) as x approaches 0 exists and is equal to f(0), which is also 0. Therefore, f(x) = |x| is continuous at x=0. To prove that f(x) = |x| is not differentiable at x=0, we need to show that the derivative of f(x) does not exist at x=0. The derivative of f(x) can be found by considering the left-hand derivative and the right-hand derivative separately. For the left-hand derivative, as x approaches 0 from the left side (x < 0), f'(x) = -1. For the right-hand derivative, as x approaches 0 from the right side (x > 0), f'(x) = 1. Since the left-hand derivative (-1) is not equal to the right-hand derivative (1), we can conclude that the derivative of f(x) does not exist at x=0. Therefore, f(x) = |x| is not differentiable at x=0.