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$f(x) = \begin{cases} -1, & \text{if $x<0$} \\[2ex] 0, & \text{if $x = 0$} \\[2ex] 1, & \text{if $x>0$} \end{cases}$

How do I prove that the limit $\lim_{x\to0}f(x)$ doesn't exist using epsilon-delta definition?

I don't know how to proceed from $|f(x) - L| <\epsilon$, since I don't know how to define what $f(x)$ is when $x$ is getting close to $0$

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  • $\begingroup$ A very brief sketch of the proof would be: "It cannot have a limit $L\ge 0$, because $f(x)=-1$ arbitrarily close to $0$, and so $|f(x)-L|\ge 1$ arbitarily close to $0$. On the other hand, it can't have a limit $L<0$ either, because $f(x)=1$ arbitrarily close to $0$..." Can you express that reasoning in epsilon-delta form? $\endgroup$
    – hmakholm left over Monica
    Dec 3, 2014 at 2:56

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Show $\lim_{x \to o^+} f(x) = 1, \lim_{x \to 0^-} f(x)=-1$.

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  • $\begingroup$ This actually will com straight from the definition and taking $\delta = \epsilon$. $\endgroup$
    – Mr.Fry
    Dec 3, 2014 at 2:57

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