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$$ \lim_{x\rightarrow 0} \frac{x^2\lfloor1/x^2\rfloor} {x^2 + 2} $$ Does this limit exist? The floor here confuses me a lot, and I could not guess whether limit exists or not.

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  • $\begingroup$ It's better to learn latex/mathjax, so that we can easily understand your question and help you. $\endgroup$ – Math Lover Dec 9 '17 at 20:40
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    $\begingroup$ Hint: $u-1\lt\lfloor u\rfloor\le u$. Let $u=1/x^2$ and see what the squeeze theorem says. $\endgroup$ – Barry Cipra Dec 9 '17 at 20:43
  • $\begingroup$ Check how other questions are posted. Not only the writing is not at a desired level so that people can help but you show no effort at all. MSE is not a site for solutions to homework questions or free tutoring lessons. You need to show at least some effort and a more detailed question. This is not your first post! $\endgroup$ – Kal S. Dec 9 '17 at 20:44
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hint

Put $t=x^2$.

For $t\ne 0$, We have

$$\frac {1}{t}-1 <\left\lfloor\frac {1}{t}\right\rfloor \le \frac {1}{t} $$

Multiply by $t $ to find that the limit is $\frac {1}{2}$.

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