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Consider $f(x)=\left\{\begin{matrix} 1-x& 0

Is $f$ continuous at $0$ (or the other endpoints)? I tried verifying it with the epsilon delta argument when the limit is possible $1$ or $0$ and I got them both to agree...

My first conjecture was that if the limit was $1$ as x goes to $0$, I would choose $\delta = \epsilon$. If the limit was $0$ as $x\to0$, then I would choose $\delta = \epsilon+1$

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$\lim_{x\to 1/2^-}f(x)=\lim_{x\to 1/2^-}(1-x)=\frac{1}{2}\neq 0= f\left(\frac{1}{2}\right)\Longrightarrow \,f\,\,\text{is not continuous at}\,\,\frac{1}{2}$

Try now to show something similar for $\,x=0\,$

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    First, there is none as the function isn't defined on the right of $\,1/2\,$, but even if it was: who cares? *The limit* exists iff both one sided limits exist, and the func. is continuous there iff the limit exists finitely and equals the function's value there, so if a one sided limit doesn't equal the function's value we're done, even if we're talking only about one sided continuity.2012-12-24
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The answer is no. This is because $\lim_{x\rightarrow 0^+}f(x)\not=f(0)$

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    Yes you are right. I should have said this2012-12-24