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.
how to show that g(t) is continuous?
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fourier-analysis
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0$g$ is not continuous at $0$ (at least not in the Euclidean topology). It is evidently continuous everywhere else. – 2012-10-25
1 Answers
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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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0That point is, however, a *removable discontinuity* ... so if you re-define the function at that one spot, it becomes continuous. – 2012-10-25