Limits problem involving greatest integer function and an unknown function. 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?
 A: 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 :)
A: 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. 
A: 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.
