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So I have the function: $$f:\left[-1,2\right]\:\bigcup \left\{3\right\}\rightarrow \mathbb{R}$$ $$f\left(x\right)\:=\:x,\:for\:x\in \left[-1,2\right]\:and\:f\left(x\right)=7,\:for\:x\:=\:3$$

And I have to prove continuity in point $x=3$ I know the function is continuous in that point, since it's an isolated point. However, I have to prove continuity using the definition. That means I must prove that for any sequence $\left(x_n\right)$ with $\lim _{n\to \infty }\left(x_n\right)=x_o$ we have $\lim _{n\to \infty }\left(f\left(x_n\right)\right)=f\left(x_o\right)$ I was thinking of having $\left(x_n\right)\in \left[-1,2\right]\:and\:\left(v_n\right)\in \left\{3\right\}$ and then proving that the limits of those two as n approaches infinity are equal and equal to $f(3)$, but that doesn't seem to work.

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  • $\begingroup$ The only sequence $x_n$ in the domain that converges to $x_o=3$ is the (eventually) constant sequence $x_n=3$ $\endgroup$ – GoodDeeds Feb 2 '16 at 14:28
  • $\begingroup$ @GoodDeeds Tail constant, to be precise. $\endgroup$ – Gregory Grant Feb 2 '16 at 14:29
  • $\begingroup$ Yes edited it, thank you. $\endgroup$ – GoodDeeds Feb 2 '16 at 14:29
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Hint:

  1. You only need to prove that for every sequence of elements from the domain of $f$.
  2. All sequences with elements from the domain of $f$ which converge to $3$ are constant from some point on.
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