0
$\begingroup$

Use limit theorems to show that the following function is continuous on $[0,1]$. $$f(x)=\begin{cases} e^{-\frac{1}{x}} & :x \neq0 \\ 0 & : \text{$x= 0$} \end{cases}$$

Here is the answer which is given to me

$*$ We know that $\space\dfrac{1}{x}$ is continuous for $x \neq 0$

$*$ Hence $\space-\dfrac{1}{x}$ is continuous on $(0,1]$ and $\space e^{-\frac{1}{x}}$ continuous on $(0,1]$

$*$ For $x=0$ we must show that, $$\lim_{x \to 0}f(x) = f(0)=0$$

$*$ Since $f(x)=e^{-\frac{1}{x}}$ for $x\neq 0$

$$\lim_{x \to 0}f(x) = \lim_{x \to 0}e^{-\frac{1}{x}}$$

$*$ Consider, $$\lim_{x \to 0^{+}}f(x) = \lim_{x \to 0^{+}}e^{-\frac{1}{x}}=0$$

$\therefore$ The limit of the function as $x \rightarrow 0^+$ is equal to the function value at $x \neq 0$

That is $$\lim_{x \to 0}f(x) = f(0)=0$$

$\therefore$ $f$ is continuous at $0$


I feel some point are wrong here because $\lim_{x \to 0^{-}}f(x)=\infty$ so we must show for $x=0$ $$\lim_{x \to 0^{+}}f(x) = f(0)=0$$ not $$\lim_{x \to 0}f(x) = f(0)=0$$ this

Is there anything wrong what I am saying please tell me?

$\endgroup$
2
  • 1
    $\begingroup$ To verifying the continuity of $f$ on $[0,1]$, it is enough to check the continuity of $f$ on $(0,1]$, which you have done in the first and second *, and to see the right continuity of $f$ at $0$, which you have done in the fifth *. The rest is superfluous. $\endgroup$
    – ARA
    Dec 31 '20 at 15:02
  • $\begingroup$ @AlirezaAhmadi thank you $\endgroup$
    – puka
    Dec 31 '20 at 15:03
0
$\begingroup$

You are wrong when you claim that $\lim_{x\to0^-}f(x)=\infty$. The domain of $f$ is $[0,1]$, and therefore it doesn't make sense to mention $\lim_{x\to0^-}f(x)$.

$\endgroup$
9
  • $\begingroup$ Sorry for the inconvenience I wanted to say $\lim_{x \to 0^{+}}f(x) = f(0)=0$ $\endgroup$
    – puka
    Dec 31 '20 at 14:48
  • $\begingroup$ It was typo now I corrected Is that correct? $\endgroup$
    – puka
    Dec 31 '20 at 14:48
  • $\begingroup$ Yes, you haved edited it, but you are still asserting that $\lim_{x\to0^-}f(x)=\infty$. $\endgroup$ Dec 31 '20 at 14:50
  • $\begingroup$ But in the answer we show $\lim_{x \to 0^{}}f(x) = f(0)=0$ is this correct? since $\lim_{x \to 0^{-}}f(x) = \infty$ $\endgroup$
    – puka
    Dec 31 '20 at 14:51
  • $\begingroup$ Yes, it is correct that $\lim_{x\to0}f(x)=f(0)=0$. And, as I have explained in my answer, the expression $\lim_{x\to0^-}f(x)$ makes no sense. $\endgroup$ Dec 31 '20 at 14:54
0
$\begingroup$

To verifying the continuity of $f$ on $[0,1]$, it is enough to check the continuity of $f$ on $(0,1]$, which you have done in the first and second $*$, and to see the right continuity of f at $0$, which you have done in the fifth $*$. The rest is superfluous.

The evaluation of $\lim_{x \to 0^{-}}f(x)=\infty$ is correct, and $f$ is not continuous on $\mathbb{R}$ and in particular at $0$. Nevertheless, it is continuous on $[0,1]$. See its graph!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.