Discontinuity of the following function.

Question

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 ?

Answer 1

hint

$$\lim_{x\to 2^-}e^{\frac {1}{x-2}}=e^{-\infty}=0$$

and at $2^+$, it is $+\infty $.

$$\lim_{x\to 3}e^{\frac {1}{(x-3)^2}}=+\infty $$

from this

$$\lim_{2^+}f (x)=\lim_3f (x)=\frac {1}{+\infty}=0$$

it is not continuous neither at $2$ nor at $3$.

Answer 2

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 $x=3$.