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Given that $\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$, find $\lim_{x \to 0} \lfloor f(x) \rfloor$ and find if $\lim_{x \to 0} \lfloor \frac{f(x)}{x} \rfloor$ exists. My math teacher says that since the denominator in the first limit is non negative and the limit itself is positive, he says that $\lim_{x \to 0} f(x) = 0^{+}$ and thus $\lim_{x \to 0} \lfloor f(x) \rfloor = 0$. I find this acceptable but my friend assumes $f(x) = 2x^{2} + \infty^{-}x^{3}$ and claims that $\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$ but that $\lim_{x \to 0} \lfloor f(x) \rfloor$ does not exist as $\lim_{x \to 0^{+}} f(x) = 0^{+}$ and $\lim_{x \to 0^{-}} f(x) = 0^{-}$. So who's right and who's wrong? If either of the two are wrong please explain why?

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  • $\begingroup$ If $f(x)=2x^2$, does it satisfy any of either of your claims? $\endgroup$ – Andrew Chin Nov 8 at 15:49
  • $\begingroup$ Your math teacher is correct. $\endgroup$ – Paramanand Singh Nov 8 at 15:58
  • $\begingroup$ Thanks for answering Andrew Chew! If $f(x) = 2x^{2}$ then quite obviously, $\lim_{x \to 0} frac{f(x)}{x^{2}} = 2$ and $\lim_{x \to 0} \lfloor f(x) \rfloor = 0$. But the problem is that $f(x)$ can be $2x^{2}$ or it can be any other expression that satisfies $\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$. We need to find what value $\lim_{x \to 0} \lfloor f(x) \rfloor$ equals irrespective of the function $f(x)$, as long as it satisfies $\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$. $\endgroup$ – K Darshan Nov 8 at 17:17
  • $\begingroup$ Paramanand Singh, I appreciate your effort but can you give the reason as to why my friend was wrong? $\endgroup$ – K Darshan Nov 8 at 17:20
  • $\begingroup$ Oh what I would give if tomorrow I'd wake up and find everyone has stopped plugging $\infty$ into expressions and treating it as though it were a number! $\endgroup$ – fleablood Nov 9 at 1:34
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The problem is, simply put, that an expression like $\infty^- \cdot x^3$ doesn't make sense; you won't find an actual function (meaning a function which only uses numbers and not $\infty$) which behaves like that. Even if you take a function like $f(x) = 2x^2 + (-10000) \cdot x^3$, as you approach zero, eventually the $x^3$ term will not matter anymore: it will be much smaller than the $2x^2$ term, if only the $x$ you insert is "small enough", so close enough to zero. That means that for $x$ small enough, this $f(x)$ will still be greater than $0$.

So yeah, your teacher is correct. In general, you should always be very skeptical when people use infinity like that. Without proper care, infinity doesn't actually make a whole lot of sense :)

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Elaborating on my comment the counter-example given by your friend does not make any sense. You can't write expressions like $f(x) =2x^2+\infty^{-}x^3$. The usage of symbol $\infty$ is always specified by specific definitions and typical examples are expressions like $x\to\infty $ and $\lim_{x\to 0}1/x^2=\infty $.

On the other hand the correct argument goes like this. Since $f(x) /x^2\to 2$ the expression $f(x) /x^2$ is positive as $x\to 0$ and hence $f(x) >0$ as $x\to 0$. Further $$f(x) =x^2\cdot \dfrac {f(x)} {x^2}\to 0\cdot 2=0$$ and hence $f(x) <1$ as $x\to 0$. It follows that $\lfloor f(x) \rfloor =0$ as $x\to 0$ and thus $\lim_{x\to 0}\lfloor f(x) \rfloor =0$.

The expression $f(x) /x= x(f(x) /x^2)$ also tends to $0$ but is positive if $x\to 0^{+}$ and negative if $x\to 0^{-}$ and hence $\lfloor f(x) /x\rfloor =0$ if $x\to 0^{+}$ and $\lfloor f(x) /x \rfloor =-1$ as $x\to 0^{-}$ so that the limit $\lim_{x\to 0}\left\lfloor \dfrac{f(x)} {x} \right\rfloor$ does not exist.

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I'm not entirely sure I understand your teacher's argument (I sure as heck don't get your friend's), but I think it is valid.

$1 > \epsilon > 0$ and there is a $\delta$ so that $|x|< \delta$ implies $|\frac {f(x)}{x^2} - 2| < \epsilon$ so $2-\epsilon \frac {f(x)}{x^2} < 2-\epsilon$. Thus $1 < \frac {f(x)}{x^2} < 3$ and $f(x) > 0$ and $x < \min(\delta,\frac 12)$ then

$0 < x^2 < \frac {f(x)} < 3x^2 < \frac 34$ so $0 <f(x) < \frac 34$ and $\lfloor f(x)\rfloor = 0$ for all $x < \min(\delta, \frac 12)$.

Which means $\lim_{x\to 0}\lfloor f(x)\rfloor = 0$.

Which I think is your teacher's argument.

I can't see why your friend assumes $f(x) = 2x^2 + \infty - x^3$, which doesn't even make sense. (What is $f(5)$? Is it $\infty -25$? What's that?) so I can't tell you why he is wrong other than that what he says makes no sense.

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  • $\begingroup$ I'm guessing here but you probably didn't understand what $0^+$, $\infty^-$, etc. actually mean. Generally a number like $a^{+}$ is a number just greater than $a$, something like $a+0.000...1$ and $a^{-}$ is a number just less than $a$, something like $a-0.000...1$. What I meant by $\infty^{-}$ is that it's a number that is very large but not $\infty$. However, on second thoughts, I believe that a number like that doesn't exist. $\endgroup$ – K Darshan Nov 9 at 13:27
  • $\begingroup$ Infinity is NOT a number. $\infty^{-}$ is oxymoronic and self-contradictory. $f(x) = 2x^2 +\infty^{-}x^3$ is meaningless and illdefined (what number is $f(5) = 50+150\infty^-$?). Your friend is more than wrong. Your friend is spouting uninterpretable gibberish. $\endgroup$ – fleablood Nov 10 at 16:10

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