Let $$f(x)=\begin{cases} 1&\text{if }x\leq 0\\ -1&\text{if }x>0 \end{cases}$$ Prove, using the precise definition of the limit, that $\lim\limits_{x\rightarrow 0} f(x)$ does not exist.
I know that this limit obviously does not exist as it does not approach the same values from the left and right hand side, but I am not sure how you would use an epsilon-delta proof to show this. Any hints are appreciated! Thanks!
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Since I don't have enough reputation to comment, I will add to Jonathan Pal's answer with another answer.
Here is Jonathan's exact answer:
What you want to show is that there exists an $ϵ>0$ where for all $δ>0$ , there are some $x,y∈(0−δ,0+δ)$ where $|f(x)−f(y)|>ϵ$ .
In this case, let $ϵ=1$ . Then for all $δ>0$ , we have that $−\frac{δ}{2},\frac{δ}{2}∈(−δ,δ)$ , $f(−\frac{δ}{2})=1$ and $f(\frac{δ}{2})=−1$ , and $|f(−\frac{δ}{2})−f(\frac{δ}{2})|=2>1=ϵ$ .
And you were wondering how $|f(−δ2)−f(δ2)|>1$ could be translated into $|f(x)−L|>1$. ($\epsilon=1$)
The fact that $|f(−δ2)−f(δ2)|>1$ actually doesn't prove the non-existence of the limit, but a little modification at the end can work.
Since $|-\frac{\delta}{2}|<\delta$, we can say that $|f(-\frac{\delta}{2})-L|<\epsilon$, or $|1-L|<1$. This means $1-L<1$, or $L>0$. By following the same process with $x=\frac{\delta}{2}$, you can get a contradiction on the value of $L$ and finish the proof.
What you want to show is that there exists an $\epsilon > 0$ where for all $\delta > 0$, there are some $x, y \in (0 - \delta, 0 + \delta)$ where $|f(x) - f(y)| > \epsilon$.
In this case, let $\epsilon = 1$. Then for all $\delta > 0$, we have that $-\frac{\delta}{2}, \frac{\delta}{2} \in (-\delta, \delta)$, $f(-\frac{\delta}{2}) = 1$ and $f(\frac{\delta}{2}) = -1$, and $|f(-\frac{\delta}{2}) - f(\frac{\delta}{2})| = 2 > 1 = \epsilon$.