<s>[INST] Proving $f(x)=\frac{1}{x}$ is continuous on the interval $(0,1)$ 
Prove $f(x)=\frac{1}{x}$ is continuous on the interval $(0,1)$.

My attempt:
$$
\left|{\frac{1}{x}-\frac{1}{c}}\right|<\epsilon\\
\left|{\frac{c-x}{xc}}\right|<\epsilon\\
\frac{\left|c-x\right|}{xc}<\epsilon \qquad\text{as $x$ and $c$ are both positive} 
$$
Suppose $|x-c|<\frac{c}{2}$ so that $\frac{c}{2}<x<\frac{3c}{2}$. Then we could obtain $\frac{1}{x}<\frac{2}{c}$
$$
\frac{1}{x}<\frac{2}{c}\implies \frac{1}{xc}<\frac{2}{c^2}\implies \frac{\left|c-x\right|}{xc}<\frac{2}{c^2}|c-x|<\epsilon.
$$
By observation, setting $|c-x|<\frac{c^2\epsilon}{2}=\delta$.
Is this proof looks right to you? Any comment and help are much appreciated!
 [/INST] It is correct for the most part. It would seem that you proved that$$|x-c|<\frac{c^2\varepsilon}2\implies\left|\frac1x-\frac1c\right|<\varepsilon,\tag1$$and that therefore $\delta=\frac{c^2\varepsilon}2$. However, you only proved $(1)$ under the assumption that $|x-c|<\frac c2$ and therefore$$\delta=\min\left\{\frac{c^2\varepsilon}2,\frac c2\right\}.$$
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