How a Absolute value Function can be continuous?

[sabyakgp]

Hello friends,

I am quite confused how an absolute function is called a continuous one. f(x) = |x| has no limit at x=0 , that is when x > 0 it has a limit +1 {+.1, +.01, +.001} and -1 when x <0 {-.1, -.01, -.001} that is the reason it's not differentiable (left and right side limits are not the same, limit does not exist). But how then it can be continuous, for one of the essential conditions for continuity is that a limit must exist for the function at the specified value (in this case x=0)? I must admit that geometrically it's indeed a continuous function, but analytically I fail understand how it is so?

I have just started learning Calculus and not very strong in Maths and I think I must have got something wrong. Can you please help me?

Best Regards,
Sabya

 

[gb7nash]

Hello friends,

I am quite confused how an absolute function is called a continuous one. f(x) = |x| has no limit at x=0 , that is when x > 0 it has a limit +1 {+.1, +.01, +.001} and -1 when x <0 {-.1, -.01, -.001}

No. f(x) = |x| is continuous at x = 0 (in fact it's continuous everywhere) The simple way of looking at it is the following:

If you approach x = 0 from the righthand side, y approaches 0. If you approach x = 0 from the lefthand side, y still approaches 0. Also, f(0) = |0| = 0. So f(x) = |x| is continuous at 0. I'm not sure where you're getting a limit of +1 and -1?

that is the reason it's not differentiable (left and right side limits are not the same, limit does not exist).

f(x) = |x| is not differentiable at x = 0, but not for the reason you listed. An easy way of looking at it is that there's a cusp at x = 0. There's no way to define a slope at this point. The more technical reason boils down to the difference quotient definition of the derivative.

 

[HallsofIvy]

Hello friends,

I am quite confused how an absolute function is called a continuous one. f(x) = |x| has no limit at x=0 , that is when x > 0 it has a limit +1 {+.1, +.01, +.001} and -1 when x <0 {-.1, -.01, -.001} that is the reason it's not differentiable (left and right side limits are not the same, limit does not exist). But how then it can be continuous, for one of the essential conditions for continuity is that a limit must exist for the function at the specified value

You seem to be thinking that this means any limit involving the function must exist. That is not the case. The limit of the function values must exist (and be equal to the value of the function there) but it does NOT follow that the limit of the difference quotient must exist.

(in this case x=0)? I must admit that geometrically it's indeed a continuous function, but analytically I fail understand how it is so?

I have just started learning Calculus and not very strong in Maths and I think I must have got something wrong. Can you please help me?

Best Regards,
Sabya

You are confusing continuity and differentiability. f(x)= |x| definitely does have a limit at x= 0. If you take x= .1, .01, .001, etc., then f(x)= |x| has values of .1, .01, .001, etc. which are clearly converging to 0. In fact for x> 0, |x|= x so $\lim_{x\to 0} |x|= \lim_{x\to 0}x$ which is, of course, 0.

If you takle x= -1., -.01, -.001, etc., then f(x)= |x| has values of .1, .01, .001, etc. which again converge to 0. In fact, for any x< 0, |x|= -x so $\lim_{x\to 0}|x|= \lim_{x\to 0}-x= -(\lim_{x\to 0}x$ which is simply -0= 0. The two one sided limits are both 0 which is the value of |0| so |x| is continuous at x= 0 (and everywhere else).

When you talk about 1 or -1, you are thinking of the difference quotient which is used to calculate the derivative, not continuity. If h> 0 then (|0+h|- |0|)/h= |h|/h= h/h= 1 so, as h goes to 0, the limit is 1. If h< 0 then (|0+h|- |0|)/h= |h|/h= -h/h= -1 so, as h goes to 0, the limit is -1. The limit of the difference quotient at 0 does not exist so |x| is not differentiable at x= 0.

By the way, notice that the denominator of the difference quotient, h, always goes to 0. A "necessary" condition that the limit exist, then, is that the numerator, f(x+h)- f(x), also go to 0 which is the same as saying the function must be continuous- but the other direction does not work. For example, in the limit
$$\lim_{x\to 2}\frac{x- 2}{(x-2)^2}$$
both numerator and denominator go to 0 but the limit does not exist.

 

[sabyakgp]

Thanks a lot gb7nash and HallsofIvy. I was quite mistaken.