Discontinuity of the following function. I want to find out the discontinuities of the function $f$ defined as
$$f(x)=\begin{cases} \dfrac{1}{1+e^{1/(x-2)}+e^{1/(x-3)^2}}, & x\neq2,x\neq3, \\[6pt] 1, & x=2, \\[6pt] \dfrac{1}{1+e}, &x=3. \end{cases}$$
My attempt :
Here we only have to check the continuity of $f$ at $x=2$ & $x=3$. As $x\to2$, $\frac{1}{x-2}\to\infty$ ,i.e., $e^{1/(x-2)}\to\infty$ ,i.e., $f(x)\to0$. But $f(2)=1\neq0$. So, $f$ is not continuous at $x=2$. Similar procedure results in discontinuity at $x=3$. Therefore the set of discontinuity of $f$ is $\lbrace2,3\rbrace$.
Am I right in the above argument ? Can someone give me an alternative solution ?

I am confident that you are right. You can also check it online for where the function is around $x=2$ and $x=3$ on derivative-calculator.net. Ignore the derivative, just focus on the graph of the function. Here is the function:
$1/(1+e^{1/(x-2)}+e^{1/(x-3)^2})$
Edit: I checked it; obviously it is not $1$ for $x=2$ and not $\frac{1}{1+e}$ for $x=3$.
