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Let $f\colon \mathbb{R} \to \mathbb{R}$ be the function

$$ f(x) = \begin{cases} \ x \exp\left(\frac{2x}{x^2-1}\right) &\text{if }x \in \mathbb{R} \setminus\{-1,1\},\\ \ 0 &\text{if }x = 1 \text{ or }x =-1. \end{cases}$$

Is this function continuous at $x=1$ and $x=-1$?

My guess:

I think that the values given to $f$ at $x=1$ and $x=-1$ are just hypothetical in case $f$ was continuous, but in reality $f$ is not continuous at these points, am I right?

But if what I just said is correct, what is the point of giving these values if f is actually not continuous at these points?

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    $\begingroup$ There is nothing hypothetical about the value of $f$ ad $x=1$: as the question says, $f(1)=0$. This allows you to evaluate continuity of $f$ at $1$ by means of deciding whether $\lim_{x\to 1}f(x)=f(1)$ or not. $\endgroup$
    – user239203
    Commented Jun 20, 2021 at 12:50
  • $\begingroup$ hmm..., what do you mean by "reality"? It is either continuous or not continuous. $\endgroup$ Commented Jun 20, 2021 at 12:51
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    $\begingroup$ Please do not post links to images. Have a look at math.stackexchange.com/help/notation and re-fromat your question accordingly. $\endgroup$
    – Gary
    Commented Jun 20, 2021 at 12:52
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    $\begingroup$ The place to start, is to stretch your intuition by using a calculator to evaluate (for example) $f(1.01)$. $\endgroup$ Commented Jun 20, 2021 at 13:09

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