Why is #f(x)=sqrtx# continuous?

1 Answer
Jim H
Apr 23, 2015

For every number #a# in #(0,oo)#, we have #lim_(xrarra)sqrtx = sqrta#. That is the definition of continuous at #a#.

At #0#, we have #lim_(xrarr0^+) sqrtx = sqrt0# which is the definition of continuous from the right at #0#.

So #sqrtx# is continuous on #[0, oo)#.

#lim_(xrarra)sqrtx = sqrta# can be proven using the definition of limit.

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