Question

PROVE.................

Answer 1

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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.

Answer 2

Step-by-step explanation:

\huge\bf\red{AnswEr:-}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.