Consider $$f(x)=\left\{\begin{matrix} 1-x& 0 Is $f$ continuous at $0$ (or the other endpoints)? I tried verifying it with the epsilon delta argument when the limit is possible $1$ or $0$ and I got them both to agree... My first conjecture was that if the limit was $1$ as x goes to $0$, I would choose $\delta = \epsilon$. If the limit was $0$ as $x\to0$, then I would choose $\delta = \epsilon+1$
Very basic continuity question
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calculus
2 Answers
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$$\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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0That's what I thought too. What happened to the right limit? – 2012-12-24
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1First, 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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0Is there even an $\lim_{x\rightarrow 0}f(x)$ – 2012-12-24
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0Yes there is. Note that the limit of f at a does not consider f(a) – 2012-12-24
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1Well, there is a *one-sided* limit at zero, not "the limit", as the function isn't defined on negative reals. – 2012-12-24
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0Yes you are right. I should have said this – 2012-12-24