# 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?

• If $f(x)=2x^2$, does it satisfy any of either of your claims? – Andrew Chin Nov 8 at 15:49
• Your math teacher is correct. – Paramanand Singh Nov 8 at 15:58
• 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 $\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$. – K Darshan Nov 8 at 17:17
• Paramanand Singh, I appreciate your effort but can you give the reason as to why my friend was wrong? – K Darshan Nov 8 at 17:20
• 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! – fleablood Nov 9 at 1:34

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 :)

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.

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 $$0 and $$\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.

• 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. – K Darshan Nov 9 at 13:27
• 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. – fleablood Nov 10 at 16:10