I need a proof of this: if f continuous at c then prove |f| continuous at c?

Question

Need rigorous proof

Answer 1

It's intuitive really.

Suppose that the limit of f(c + δ) as δ approaches 0- and 0+ are L1 and L2 respectively.
If f(x) is continuous at c, then L1 = L2
|f(c+)| = |L2|
|f(c-)| = |L1|

Since L1 = L2
|L1| = |L2|
thus |f(c+)| = |f(c-)|
Thus |f(c)| is continuous.

Answer 2

Consider that per the triangle inequality, |x| = |x-y+y| ≤ |x-y| + |y|
|x|-|y| ≤ |x-y|. And similarly, |y| = |y-x+x| ≤ |y-x|+|x|, so |y|-|x| ≤ |y-x|. But since |y-x| = |x-y|, this means that both |x|-|y| and |y|-|x| = -(|x|-|y|) are less than or equal to |y-x|, so ||x|-|y|| ≤ |x-y|. Now we proceed with the proof:

To show that |f| is continuous at c, we need to show that ∀ε>0 ∃δ>0 such that |x-c| < δ → ||f(x)|-|f(c)|| < ε. But since since f is continuous, this means that for any given ε, there exists δ such that |x-c| < δ → |f(x)-f(c)| < ε, and since ||f(x)|-|f(c)|| ≤ |f(x)-f(c)|, this same δ also suffices to ensure ||f(x)|-|f(c)|| < ε. So we are done.