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A function which is continuous at only one point.

It is

displaymath159

This function is continuous only at the point x=0. Why? Well, it's certainly not continuous at any other point, since if tex2html_wrap_inline163 , then by taking a sequence of rational numbers converging to tex2html_wrap_inline165 and then a sequence of irrational numbers converging to tex2html_wrap_inline165 , you can see that tex2html_wrap_inline169 doesn't exist. But why is f continuous at 0? We know that f(0)=0 from the definition. We also have that tex2html_wrap_inline175 for all x (since f is always either x or -x). Since tex2html_wrap_inline185 , it follows from the pinching (or sandwich) theorem that tex2html_wrap_inline187 . Since this is f(0), this means that f is continuous at 0.



Tom Vogel
Mon May 5 12:53:33 CDT 1997