Proving discontinuity at $\,x =1\,$ using $\,\varepsilon$-$\delta\,$ definition I was doing a proof for discontinuity using $\,\varepsilon$-$\delta\,$ definition but I’m not sure whether the proof is right. Would you mind checking it for me please, thanks!
$g(x)=\begin{cases}\dfrac{1}{1-x}\;,\quad& x\neq1\\0\;,&x=1\end{cases}$
For the proof, I have chosen $\,\varepsilon = 1$.
And said for $x\in [0,2]\cap(1-\delta, 1+\delta)\,$ where $\,x\neq1$
$|x-1|<\delta\,$ but $\,|g(x)-g(1)|\geqslant1\,,$
Is this right?
Thanks!

The function g(x) is continuous at $x=x_0$ if
$$\forall \epsilon >0, \exists \delta >0:|x-x_0|<\delta\Rightarrow |g(x)-g(x_0)|<\epsilon$$
The function g(x) is discontinuous at $x=x_0$ if
$$\exists \epsilon >0\,,\forall \delta >0\,,\exists x\in(x_0-\delta,x_0+\delta) :|g(x)-g(x_0)|\ge \epsilon\,.$$
In this case $x_0=1; g(x_0)=0\,.$
Lets take $\epsilon=1$. Observe that, for $0<|x-1|<1$,
$$
\frac{1}{|x-1|}>\epsilon,\quad\text{ or equivalently, }\quad
\left|\frac{1}{x-1}-0\right| \gt \epsilon\,.
$$
That is:
$$
|g(x)-g(1)| \gt \epsilon\,.
$$
This shows that for $\delta=1$ every $x\in(1-\delta,1)\cup(1,1+\delta)=(0,1)\cup(1,2)$ satisfies
$$\tag{1}
|g(x)-g(1)|>\epsilon.
$$
Clearly, for any other $\delta>0$ we can find at least one
$x\in(1-\delta,1)\cup(1,1+\delta)$ that satisfies (1) as well because the
intersection of this set with the previous one, $(0,1)\cup(1,2)\,,$ is not empty.
This finishes the proof that $g(x)$ is discontinuous at $x=1\,.$
