Question

1. Prove that f(x) = |x| is a continuous function.​

Answer 1

For all the x0>0 you have in a neighbourhood of x0 that |x|=x and you know f(x)=x is continuous over R. The same goes for x0<0. Then you only have to prove the continuity in 0. Considering that:

-limx→0−|x|=limx→0−−x=0

-limx→0+|x|=limx→0+x=0

you can conclude that

limx→0|x|=0

therefore the function is continuous in 0.

to prove that a function is continuous you just have to prove that it is continuous in every point (in fact, that's the definition)

Answer 2

Answer:

To show that f(x)=|x| is continuous at 0, show that limx→0|x|=|0|=0.

Use ε−δ if required, or use the piecewise definition of absolute value.

f(x)=|x|={xifx≥0−xifx<0

So, limx→0+|x|=limx→0+x=0

and limx→0−|x|=limx→0−(−x)=0.

Therefore,

limx→0|x|=0 which is, of course equal to f(0).

To show that f(x)=|x| is not differentiable, show that

f'(0)=limh→0f(0+h)−f(0)h does not exists.

Observe that

limh→0|0+h|−|0|h=limh→0|h|h

But |h|h={1ifh>0−1ifh<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.

Hope this is helpful ❤

Follow mee