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Limits & derivative of absolute value of x

  1. Feb 16, 2012 #1
    For absolute value of x, f(x) = lxl


    To prove continuous, I can express it as a piecewise function of x if x>0 and -x if x<0 and then I can take the limit of each as x approaches 0 from the negative and positive sides. Both limits give zero, and f(0) gives zero, so continuity is proven, right?

    Also, if I take the limit as h approaches 0 and get the general derivative function of x/lxl , whatever number I plug into that derivative function is going to be the slope of the tangent at a specific point. So if I pick x = 5 so that f'(5) = 1, so the slope is 1, and if I pick -5, the slope will be -1.

    However, I'm confused at x = 0.

    for the original function lxl, there is no denominator so no vertical asymptote. But for the derivative function x/lxl there is a denominator, and the denominator can't be zero. That means there must be a vertical asymptote at x = 0 on the function x/lxl.

    Now, if I plug 0 into the derivative, f'(0) = 0/l0l, I get 0/0 which is no solution.

    So which condition, specifically, determines that the function is not differentiable at x = 0? The fact that the denominator of f prime can't be zero, or that f prime of zero produces no solution?
     
  2. jcsd
  3. Feb 16, 2012 #2

    Bacle2

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    Note that the derivative is a limit as h-->0 , and this means as h-->0 either from the left , or from the right. If the limit exists, it is the same in both cases.
     
  4. Feb 16, 2012 #3
    okay, so when i take the limit as h-->0, I get a general derivative formula. And for absolute value of x, i get x/lxl.

    I think that is making good sense to me at the moment. however, where i'm confused is proving that absolute value of x is not differentiable at 0. - as per my question: ".....which condition, specifically, determines that the function lxl is not differentiable at x = 0? The fact that the denominator of f prime can't be zero, or that f prime of zero produces no solution?"

    EDIT:

    I'm still not sure if you were replying to my specific question, but after reading your response and thinking about my confusion, your note helped me in the right direction i think. So i'm going try to turn x/lxl into a piecewise function and take the limit from both sides and see what happens. i think i'm going to get 1 and -1 and then that will be the proof that it's not differentiable.


    EDIT 2:

    Okay, so i just turned f prime into a piecewise function and took the limit as x approaches zero from both sides. I chose limit of 0/0 for x>0 and 0/-0 for x<0, and both show no solution. Maybe i'm also confused on how to show continuity of x/x ?
     
    Last edited: Feb 16, 2012
  5. Feb 16, 2012 #4

    Bacle2

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    So the idea is taking the limit x/|x| as x->0 : Lim_>h->0 [|0+h|-|0|]/h

    Then |x| is differentiable at 0 iff (def.) the limit exists.
     
  6. Feb 16, 2012 #5

    Deveno

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    have you tried graphing x/|x|? what does it look like?

    can you think of any value you could assign for x = 0 that would make x/|x| continuous?

    surely lim h→0+[(x+h) - x]/h = 1, while

    lim h→0-[-(x+h) - (-x)]/h = -1.

    the moral of this story is: the definition of the derivative. learn it, love it, live it.

    intuitive, and totally not formal aside: derivatives don't like "pointy places".
     
  7. Feb 16, 2012 #6

    Okay, i'm a little confused. I know taking the limit x/lxl as h->0 of f(x+h) - f(x)/h would just give me a third derivative of lxl. So why does taking the limit of x/lxl need to be set up the same way as taking the derivative of x/lxl?
     
  8. Feb 16, 2012 #7
    Yes i graphed it. It's like a step function and not continuous at 0.

    But i'm confused by the limit that you are taking. It looks like the same kind of limit you would take to get a second derivative. I don't need to find the second derivative right?

    The only way to find the limit x/lxl as x->0 from both sides is to create a piecewise function, right? I know that if I create a piecewise function and chose x = 1 for x>0 and chose x = -1 for x<0, then I will get 1 and -1. But I don't know if that means I am proving discontinuity, thus nondifferentiability of lxl. And I also don't know if what I just described is corresponding to the limit that you are showing, limit as h approaches 0.
     
    Last edited: Feb 16, 2012
  9. Feb 16, 2012 #8

    HallsofIvy

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    If x> 0, |x|= x so
    [tex]\lim_{h\to 0^+} \frac{|h|- 0|}{h}= \lim_{h\to 0}\frac{h}{h}= 1[/tex]

    If x< 0, |x|= -x so
    [tex]\lim_{h\to 0^-} \frac{|h|- 0|}{h}= \lim_{h\to 0}\frac{-h}{h}= -1[/tex]

    Again, for the limit to exist, so that the derivative at x=0 exist, those two one sided limits would have to be the same.
     
  10. Feb 16, 2012 #9

    Deveno

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    to prove |x| is not differentiable at 0, you have to use the definition.

    |x| is an abbreviation for the function:

    f(x) = -x , x < 0
    f(x) = x, x ≥ 0.

    it's pretty clear that the two "pieces" have constant derivative, which is well-defined everywhere except 0:

    f'(x) = -1, x < 0
    f'(0) = ??????
    f'(x) = 1, x > 0

    now to find the derivative AT 0, the two (one-sided) limits HallsofIvy said in his last post:

    a) have to exist
    b) have to be equal

    (a limit cannot have two separate values at one point).

    we sail past (a) just fine, but (b) is the killer. the two (one-sided) limits are unequal, so the derivative DOES NOT EXIST (at 0).

    people often get confused by what a derivative IS.

    for example, people will say something like: "the derivative of x2 is 2x". that is all wrong. the derivative is a LIMIT. now for certain "nice" functions (these are called smooth functions, if you must know), the derivative of f can often be stated with a pretty little formula....but these formulas are NOT what the derivative is, the derivative is a limit, which just happens to have a nice form for some nice functions.

    and trust me when i say this, there are lots of evil functions out there, waiting for you to blindly rush into their traps. discontinuities, unbounded values, corners, and wild oscillations, just to name a few of the bizzare ways functions can thwart our atttempts to differentiate them.
     
  11. Feb 16, 2012 #10

    Bacle2

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    Sorry, my bad. I meant find the limit as h->0 of |h|/h and check to see if the limit
    exists, meaning the two-sided limit.
     
  12. Feb 16, 2012 #11
    [quote="Wikipedia at
    http://en.wikipedia.org/wiki/Derivative] [Broken]

    .
    .
    .
    derivative as a function

    Let ƒ be a function that has a derivative at every point a in the domain of ƒ.
    Because every point a has a derivative, there is a function that sends the
    point a to the derivative of ƒ at a.
    [itex] > > > [/itex]This function is written f′(x) and is called the
    derivative function or the derivative of ƒ. [itex] < < < [/itex]
    The derivative of ƒ collects all the derivatives of ƒ at all the points in
    the domain of ƒ.

    .
    .
    .

    f functions: The derivative is an operator whose domain is the set of all
    functions that have derivatives at every point of their domain and whose
    range is a set of functions. If we denote this operator by D, then D(ƒ) is
    the function f′(x). Since D(ƒ) is a function, it can be evaluated at a
    point a. By the definition of the derivative function, D(ƒ)(a) = f′(a).

    [/quote]

    I don't know what kind of distinction you're trying to make.

    [itex]If \ \ the \ \ function \ \ is \ \ x^2, \ \ then \ \ the \ \ derivative \ \ function [/itex]

    [itex]is \ \ f'(x) = 2x. \ \ \ Or, \ \ f'(x) = 2x \ \ is \ \ called \ \ the \ \ derivative \ \ of \ \ f[/itex]

    [itex](where \ \ here \ \ f \ \ is \ \ the \ \ function \ \ f(x) = x^2).[/itex]
     
    Last edited by a moderator: May 5, 2017
  13. Feb 16, 2012 #12

    Deveno

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    for "differentiable-everywhere" functions (functions for which the limit that is the derivative exists at all elements of the domain of definition of f, typically all of R, if no specific domain is given), we can define the derivative as the function which has as it's value at x, f'(x).

    but "differentiable everywhere" is a rather strong condition. to know that a function is "differentiable everywhere" presumably, you've already taken it out for dinner a couple of times, maybe even bought it a gift or two. it's definitely NOT the first date (i suppose previously dating a function's sister, or maybe even second cousin, might lead you to make certain presumptions...)

    my point is, a lot of people apply the differentiation formulas FIRST, and then check to see if that's kosher. that's backwards. you can't apply the formula unless you know FIRST, that the function is indeed differentiable. and there's really only one way to do that: check that the derivative EXISTS.

    for example, if f(x) is a polynomial, one uses the linearity of the derivative. a close examination of the proof of linearity will reveal statements like:

    IF f AND g are differentiable (means: need to verify this first!), THEN f+g is also differentiable with (f+g)' = f'+ g'.

    the preliminaries ARE important, and the function at hand shows why: it is simply WRONG to say that:

    (|x|)' = x/|x|, because it's not true "everywhere".

    i can easily define any number of continuous functions with several points where differentiabilty may, or may not, be defined. use a sawtooth wave, and multiply it by a trig function.

    i surely do not mean to imply that f(x) = 2x isn't the function whose value at x is g'(x), where g(x) = x2. what i'm trying to say is: don't just blindly apply the power rule, every time you see an exponent. take f(x) = x1/3, for example. not differentiable (everywhere). the exceptions matter.
     
  14. Feb 16, 2012 #13
    I get it now. I was able to find where f(x) = l3x-2l is not differentiable and correctly prove that while continuous, not differentiable at its root.

    Thanks for showing me the way.
     
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