Explanation regarding continuity in piecewise function

Question

Let $ \begin{equation} f(x) = \begin{cases} e^{-1/x}, & x > 0,\\ 0, & x \le 0 \end{cases} \end{equation}$

I have proven that $f(x)$ is continuous at $x \in R\setminus {0}$, and in order to prove that it is continuous at $x=0$, consider $(x_{n}) \in \mathbb R_{+}$ as well as $(y_{n}) \in \mathbb R_{-}$

For a sequence $x_{n} \to 0, n \to \infty$, we have $f(x_{n})=e^{-1/x_{n}}$. My intuition would be to just say

$\lim_{n \rightarrow \infty} f(x_{n}) = \lim_{n \rightarrow \infty} \frac {1}{e^\frac{-1}{x_{n}}}$=$\frac{1} {e^\lim {n \rightarrow \infty}{\frac{-1}{x_{n}}}}=0$. But would this not be 'illegal' as taking the limit into the equation already constitutes it being continuous at $0$. Isn't that exactly what I am attempting to prove?

I'm struggling to understand this nuance. Any help would be greatly appreciated

Answer 1

You can justify that $\lim_{x\rightarrow 0^{-}}f(x)=\lim_{x\rightarrow 0^{+}}f(x)=f(0)$.

But $\lim_{x\rightarrow 0^{-}}f(x)=0=f(0)$ is clear.

Now we need only that $\lim_{x\rightarrow 0^{+}}f(x)=0$.

For the sequential characterisation for the right limit, one may assume $x_{n}>0$ and $x_{n}\rightarrow 0$, then the rest goes through.

Having assume $x_{n}>0$ and $x_{n}\rightarrow 0$, the only thing to deal with is to justify that $e^{-1/x_{n}}\rightarrow 0$.

This is by the spirit of composition. One may use the fact that $e^{u}\rightarrow 0$ as $u\rightarrow-\infty$. Given $\epsilon>0$, then there is some $M<0$ such that for all $u$ with $u<M$, then $e^{u}<\epsilon$.

Concentrating with this $-1/M>0$, since $x_{n}\rightarrow 0$, there exists some $N$ such that $n\geq N$ implies that $x_{n}<-1/M$, keeping in mind that $x_{n}>0$, then we are allowed to flip over to get $x_{n}^{-1}>-M$, and hence $-x_{n}^{-1}<M$.

Such an $-x_{n}^{-1}$ can be a candidate for such a $u$ just mentioned, so $e^{-1/x_{n}}<\epsilon$, this shows that $e^{-1/x_{n}}\rightarrow 0$, as expected.

Answer 2

just a hint

If $x_n>0$ and $x_n\to 0$ then

$$\frac {1}{x_n}\to +\infty $$ and

$$e^{\frac {-1}{x_n}}\to e^{-\infty}=0$$

thus $$\lim_{x\to 0^+}f (x)=0=f (0) .$$

Answer 3

What you've done is not illegal. You did not take the limit of your function, you've taken the limit of the function $f(x)=e^{-\frac{1}{x}}$, which happens to coincide with your function on $(0, \infty)$. Your function and $e^{-\frac{1}{x}}$ have the same values, therefore they have the same limit when $x\to 0^{-}$.

(Of course, you now need to prove that $\lim_{x\to 0^{-}}e^{-\frac{1}{x}}=0$, which is more-or-less done in your question; I am not sure if your write-up is rigorous enough, but this is another matter...)