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$$f(x)=\begin{cases}x^2-3 & x<1 \\ x-1& x\geq 1\end{cases}$$

Prove $f$ is not continuous at $x=1$.

$$\text{Let } \epsilon =1 , \delta>0.$$ $$\text{Choose}~x~ \text{such that}~ |x-1| < \delta \text{ and } 0<x<1.$$ $$\text{Then } |x^2-3| > |x-3|>|1-3|=2>\epsilon.$$

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    $\begingroup$ Have you said anywhere that $f(1)=0$? $\endgroup$
    – Henry
    Sep 23 at 9:34
  • $\begingroup$ The inequalities in the last line require an explanation. $\endgroup$ Sep 23 at 9:36
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    $\begingroup$ The proof is correct but in the last part is better to write$ |x^{2}-3|=3-x^{2}$ and since $-x^{2}>-1$ we get $|x^{2}-3|>2>1=\epsilon$ $\endgroup$ Sep 23 at 9:40

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