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how to show that g(t) is continuous? $g(t) = piecewise(t \neq 0, 0, t=0, 1)$ $\lim_{t \rightarrow 0} g(t) =1$? So it should be continuous at $t=1$? Well it seems this question doesn't meet you guality standards.

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$g$ is not continuous at $0$ (at least not in the Euclidean topology). It is evidently continuous everywhere else. – Lord_Farin Oct 25 '12 at 8:36
up vote 1 down vote accepted

A function $f$ is continuous at a point $a$ iff $\displaystyle\lim_a f=f(a)$.

In your case, where $g(t)=0$ for all $t\ne 0$ and $g(0)=1$, we have $\displaystyle\lim_0 g =0\ne g(1)$, so it is not continuous at $0$. We also have $\displaystyle\lim_a g=0=g(a)$ for all $0\ne a\in \Bbb R$, hence contiunous everywhere else.

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That point is, however, a removable discontinuity ... so if you re-define the function at that one spot, it becomes continuous. – GEdgar Oct 25 '12 at 11:48

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