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

1 Answer
Mar 22, 2016

See the explanation, below.

Explanation:

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.