1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Derivative at a possible corner

  1. Oct 9, 2013 #1


    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    http://i.minus.com/j1DgeK5VMMfHn.png [Broken]

    2. Relevant equations

    To simplify a limit with a square root multiply by the conjugate.

    3. The attempt at a solution

    The limit doesn't seem to exist as the left and right hand limits diverge.

    The left hand limit with f(x) = x^2 + 4 results in x^2/x = x and the limit of x as x approaches 0 from the left is 0.

    The right hand limit, after multiplying it by its conjugate, simplifies to 1/[(sqrt(x+16)) + 4]. Taking the limit of this as x approaches 0 results in 1/8.

    The limit definition of the derivative for this particular f(x) doesn't seem to exist.

    Does this imply the function is not differentiable at x = 0? I graphed it out and it seems there is a corner at x = 0. I tried looking up the mathematical definition of a corner but the best I got was that at a corner the derivative changes abruptly; it's not "smooth." The derivative does jump from being negative on the left with x^2 + 4 and to being positive with the sqrt(x+16). So what's the mathematical definition of a corner? I know for a cusp the mathematical definition is that the left and right hand limits go to infinities of different signs at the point of the cusp. So what's a corner? And am I correct that this function is not differentiable at x = 0?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 10, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper

    Be careful with the language here. The word diverge - in "f diverges as ##x \to a##", is usually used to mean the formal equivalent of "##\lim_{x \to a} f(x) = \infty##". What you mean is that the left and right hand limits are not equal, and therefore the limit does not exist. There is no divergence though.

    That's right. Note that these one-sided limits are actually the derivative of the pieces evaluated at x = 0: ##d/dx(x^2 + 4)\vert_{x = 0} = 0## and ##d/dx( \sqrt{x + 16} )\vert_{x = 0} = 1/8##. Of course, you noticed that but you were asked to use the limit definition.

    The definition is fine, the limit just doesn't exist :-)

    I'm not sure "corner" has an official definition in the context of function graphs (probably it is defined when talking about e.g. polygons). You could define it as a point where the function is continuous but not differentiable. I don't know whether "cusp" is formally defined either, but I would expect it to be similar - in any case I don't think it's necessary for any limit to go to infinity. E.g. I would say f(x) = |x| has a cusp or corner at x = 0, but I wouldn't say that f(x) = 1/x does, while only the latter involves "infinities" - that is, one-sided limits that do not exist.
  4. Oct 10, 2013 #3


    User Avatar
    Gold Member

    Thank you for the thoughtful reply! I agree that precise wording is an imperative.

    I found a definition for a cusp on the University of Chicago's website.

    http://math.uchicago.edu/~vipul/teaching-1011/152/concaveinflectioncusptangentasymptote.pdf [Broken]

    http://i.minus.com/jZ459eGBlUzbv.png [Broken]
    Last edited by a moderator: May 6, 2017
  5. Oct 10, 2013 #4
    Re cusp as per your definition not ##x^{2/3}## which has no derivatives going off to infinity at any finite point. But 1/x is an example.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted