Derivative at a possible corner

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The discussion centers on the limit and differentiability of a function at x = 0, where the left-hand limit approaches 0 and the right-hand limit approaches 1/8, indicating that the overall limit does not exist. This divergence suggests that the function is not differentiable at x = 0, which is characterized as a corner due to the abrupt change in the derivative from negative to positive. While the term "corner" lacks a formal definition in this context, it is generally understood to describe points where a function is continuous but not differentiable. The conversation also touches on the definitions of cusps and corners, noting that both involve discontinuities in the derivative without necessarily going to infinity. Ultimately, the function's behavior at x = 0 illustrates the concept of non-differentiability at a corner.
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Homework Statement



http://i.minus.com/j1DgeK5VMMfHn.png

Homework Equations



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

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

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.

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

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.
 
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

http://i.minus.com/jZ459eGBlUzbv.png
 
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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