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I mean to say how to find the limit $\displaystyle\lim_{x \to 0} \frac{[x^2]}{x^2}$? where $[x]$ denotes the greatest integer function of $x$.

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    Isn't the answer just 0. Am I overthinking?! – 2017-01-02

1 Answers 1

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For $|x|<1$, we know that $[x^2]=0$. Thus, we have

$$\frac{[x^2]}{x^2}=0\text{ when }|x|<1,x\ne0$$

So the limit is obviously $0$.

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    how can we say that |x|<1 implies [x^2]=1? – 2017-01-02
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    @s.panja No, it is equal to $0$. Clearly, since $[0.1^2]=0$, $[0.9^2]=0$, etc. – 2017-01-02
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    Nice answer! (+1) – 2017-01-02
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    but what about the value of the function in a nbd of 0? – 2017-01-02
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    @s.panja Identitically $0$, always. – 2017-01-02
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    yah thanks@simple art – 2017-01-02
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    @s.panja No problem, glad to clear up the problems! – 2017-01-02
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    @SimpleArt: so it is that tricky situation when both left and right limits at $x=a$ exist, but $f(a)$ does not – 2017-01-02
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    @Alex Clearly division by $0$ has that yes – 2017-01-02