Discontinuity of floor function
You over complicate things. Simply take the left hand limit and the right hamd limit seperately. You will find the left hand limit to be -1 and the right hand limit to be 0. Therefore, there is jump discontinuity at x=0.
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otupygak
Updated on November 23, 2020Comments
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otupygak almost 3 years
I am still getting confused with showing discontinuity of functions, here is my attempt at a question , if someone could check to see if I am going about it in the correct way.
$f: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x)=[x]$ is discontinuous at 0.
My attempt;
I decided against using the epsilon-delta definition and decided to go with the limit approach.
We want to show that $$\lim_{x \to 0}[x] \neq f(0)=0$$
Now suppose it is, then $\forall \epsilon>0 \exists \delta >0$ s.t $|f(x)-f(c)|<\epsilon $whenever $|x-c|<\delta$.
Take $\epsilon =\dfrac{1}{2}$ ; for all $x<0$ $f(x)<0$ so if $f(x)<0$ using our chosen delta we want $|f(x)-f(0)|=|f(x)|<\dfrac{1}{2}$ but this is impossible , as there exists no $\delta$ that satisfies $|f(x)|<\dfrac{1}{2}$ whenever $|x-c|<\delta$Is this correct??
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Siddharth Venkatesh over 9 yearsLooks ok. You could probably make it clearer but the argument looks correct. You don't really need to use a contradiction. Just pick $\epsilon < 1$ and then do what you did to show that no suitable $\delta$ exists.
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Tom Collinge over 9 yearsOr, just observe that $\lim_{x\rightarrow-0} f(x) = -1$ whereas $f(0) = 0$, i.e. limit as x approaches zero from below is not equal to $f(0)$, therefore not continuous at zero.
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otupygak over 9 yearsThank you very much for your answers, I just wanted to practice doing it fully,
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