Why is the function abs(x) not differentiable at x=0?

[Bassalisk]

Hello,

I am currently trying to understand some limits. |x| is not differentiable because left and right limits are different.

I also learned that |x|/x=sgn(x) (signum)

But

$$\lim_{\substack{x\rightarrow 0}} sgn(x)$$

is not defined, ergo function abs(x) is not differentiable in 0.

But isn't sgn(x) defined to be 0 if x=0, or am I misinterpreting the term "approaching".
Another example:

let

$$g(y) = \begin{Bmatrix} <br /> 1, & y \not= 0 \\ 0, & y=0 \end{Bmatrix}$$

and f(x)=0 for all x.

$\lim_{\substack{y\rightarrow 0}} g(y)=1$

$\lim_{\substack{x\rightarrow 0}} f(x)=0$

but

$\lim_{\substack{x\rightarrow 0}} (g \circ f) (x) =0$

This is example where composition of functions doesn't work as they should.

I don't understand how this composition of functions, doesn't work. Can anybody explain to me how this works? ( I found the example on the internet, but not as good explanation)

 

[mathman]

For the first question: lim sgn(x), = 1 for x-> 0 and x > 0, = -1 for x ->0 and x < 0. Therefore the limit does not exist. Defining sgn(0)=0 or anything else doesn't matter.

For the second [I'm assuming you are asking about g(f(x))]: f(x)=0 for all x, therefore this function is g(0)=0 for all x. I am not quite sure what "doesn't work".

 

[Bassalisk]

For the first question: lim sgn(x), = 1 for x-> 0 and x > 0, = -1 for x ->0 and x < 0. Therefore the limit does not exist. Defining sgn(0)=0 or anything else doesn't matter.

For the second [I'm assuming you are asking about g(f(x))]: f(x)=0 for all x, therefore this function is g(0)=0 for all x. I am not quite sure what "doesn't work".

It doesn't work like this:

$$\lim_{\substack{x\rightarrow a}} f(x)=b$$
$$\lim_{\substack{y\rightarrow b}} g(y)=c$$

$$\lim_{\substack{x\rightarrow a}} (g \circ f) (x) =c$$This feels right to me, But i cannot understand the problem above.(2nd one)

Does that "for all x" makes everything in composed function 0?

 

[mathman]

The problem (why it doesn't work) is simply the discontinuity at x=0 for g. For the composite function the argument for g is always 0. The general property for the limit involving g assumes -> 0 through non-zero values of the argument.

 

[HallsofIvy]

sgn(x)= 1 if x> 0, sgn(0)= 0, and sgn(x)= -1 if x< 0.

In particular,
$$\lim_{x\to 0^+} sgn(x)= \lim_{x\to 0} 1= 1$$
and
$$\lim_{x\to 0^-} sgn(x)= \lim_{x\to 0} -1= -1$$

Of course, $\lim_{x\to a} f(x)= L$ if and only if both $\lim_{x\to a^+} f(x)= L$ and $\lim_{x\to a^-} f(x)= L$. If the two sided limits are not the same, then the limit itself cannot exist.

As for your second problem, in general [itex]\lim_{x\to a} g(f(x))= g(\lim_{x\to a}f(x))[/tex] if and only if g is <b>continuous</b> at $\lim_{x\to a}f(x)$. In fact, that can be used as a <b>definition</b> of "continuous". Here, you function g is not continuous at x= 0.<br /> <br /> That's why the limit at 0 does not exist. It has nothing to do with the value of the function at x= 0.[/itex]

 

[Bassalisk]

sgn(x)= 1 if x> 0, sgn(0)= 0, and sgn(x)= -1 if x< 0.

In particular,
$$\lim_{x\to 0^+} sgn(x)= \lim_{x\to 0} 1= 1$$
and
$$\lim_{x\to 0^-} sgn(x)= \lim_{x\to 0} -1= -1$$

Of course, $\lim_{x\to a} f(x)= L$ if and only if both $\lim_{x\to a^+} f(x)= L$ and $\lim_{x\to a^-} f(x)= L$. If the two sided limits are not the same, then the limit itself cannot exist.

As for your second problem, in general $\lim_{x\to a} g(f(x))= g(\lim_{x\to a}f(x))$ if and only if g is continuous at $\lim_{x\to a}f(x)$. In fact, that can be used as a definition of "continuous". Here, you function g is not continuous at x= 0.

That's why the limit at 0 does not exist. It has nothing to do with the value of the function at x= 0.

My brain is swelling of information i got recently. I misplaced the definition of "not continuous". Thank you.