Are Functions Continuous at a Cusp or Corner?

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Functions can be continuous at a cusp or corner, even though they are not differentiable at those points. Continuity requires that the limit as x approaches the point exists, the function is defined at that point, and the limit equals the function's value at that point. An example is the function f(x) = |x|, which has a cusp at x = 0, is defined there, and is continuous for all x. Therefore, it is possible to have a continuous function at a cusp or corner.
Rozy
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I'm just working through some differentiability questions and have a quick question - are functions continuous at a cusp or corner? I know that functions are not differentiable at cusps or corners because you cannot draw a unique tangent at these points, but I'm not sure about continuity. From my understanding, a function is continuous if at point a if

1. The limit as x approaches a exists
2. f(a) exists or is defined
3. the limit of x approaching a = f(a)

Can you define the function at a cusp or corner and thus have a continuous function?

Thanks!
Rozy
 
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Yes, you can.
 
On a side-note, you can construct functions that are everywhere continuous, but nowhere differentiable..
 
Thanks

Thanks for responding! :smile:
 
For example f(x)= |x| has a "cusp" or corner at x= 0. Of course, f(0)= |0|= 0 so the function is certainly define there- and continuous for all x.
 

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