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

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Answer 1 · 2016-03-22T23:06:13

See the explanation, below.

To show that $f (x) = | x |$ $f(x)=absx$ is continuous at $0$ $0$ , 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 $x = 0$ .