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Show by $\varepsilon-\delta$-criterion that for each $c\in\mathbb{R}$, the function $f\colon\mathbb{R}\to\mathbb{R}$, $$ f(x)=\begin{cases}\frac{1}{x}, & x\neq 0\\c, & x=0\end{cases} $$ is not continuous in $x=0$.

My idea is to show this by contradiction. Let $c\in\mathbb{R}$ be such that the function is continuous in $x=0$. Let $\varepsilon >0$, then there exists some $\delta >0$ s.t. $$ \lvert \frac{1}{x}-c\rvert\leq\varepsilon \text{ for all }x\in\mathbb{R}\text{ s.t. }\lvert x\rvert\leq\delta. $$

Dont know how to argue to see that we have a contradiction.

My idea: Since the statement has to hold for all x in $\mathbb{R}$ with $\lvert x\rvert\leq\delta$, we can choose $x$ very small. But then $1/x$ tends to $\pm\infty$ and hence $\lvert \frac{1}{x}-c\rvert$ will not be $\leq\varepsilon$.

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  • $\begingroup$ The statement has to hold for at least one $|x| \le \delta$, not all. $\endgroup$ – copper.hat May 30 '16 at 17:45
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Suppose $x\neq 0$. Note that $|{1 \over x } -c | \ge {1 \over |x|} - |c|$.

In this case, we can try to find $x$ such that ${1 \over |x|} - |c| > 1$, rearranging gives $|x| < {1 \over 1 + |c|}$.

Hence if $x \in (-{1 \over 1 + |c|}, {1 \over 1 + |c|})$ then $|{1 \over x } -c | > 1$.

Aside: You need to show that there exists some $\epsilon>0$ such that for all $\delta>0$, there exists some $x$ such that $|x| < \delta$ and $|f(x)-f(0)| > \epsilon$.

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Take the succession ${x_n}=1/n$. If $f$ is continuous in zero, then $f(x_n)\rightarrow c$ when $n\rightarrow\infty$, but this isn't true.

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