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Derivative at a possible corner
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[QUOTE="CompuChip, post: 4532778, member: 81086"] 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. [/QUOTE]
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Derivative at a possible corner
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