1
$\begingroup$

I'm currently taking introduction to Calculus and I've been presented with this limit involving the greatest integer function (GIF):

$$\lim_{x \to 2^-} \frac{\lfloor x \rfloor - 1}{\lfloor x \rfloor - x}$$

Now since $x \to 2^-$ I figured I could immediately evaluate the limits of the first terms of the numerator and denominator and replace it with 1. This makes the numerator 0 and the denominator will be $1 - x$. After evaluating the limit of the denominator I arrive at $\frac{0}{-1}$ which is basically 0. However looking at the problem's answer keys it said that the limit should have been $-\infty$. I wasn't sure where I got it wrong so I thought that the key must be wrong for that item until I tried solving for this next limit:

$$\lim_{x \to 1^+} \frac{\lfloor x^2 \rfloor - \lfloor x \rfloor^2}{x^2 - 1}$$

Since $x \to 1^+$, I thought that $\lfloor x^2 \rfloor$ and $\lfloor x \rfloor$ would both resolve to 1 right? And that's what I did but the limit of the entire function then becomes an indeterminate form of type $\frac{0}{0}$. Now I know that I should rewrite the function in order to get rid of the terms that would cause it to become $\frac{0}{0}$ and factoring the denominator gives me $(x + 1)(x - 1)$ which will become $(2)(0^+)$ but given that the numerator is 0, isn't that the same indeterminate form? The answer key says that the limit is 0.

Am I missing something here? Is there a way to manipulate or rewrite these GIF that simply hasn't been taught to us?

EDIT

If you guys think that the question or the answer key must be wrong then please feel free to tell me :) I'm preparing for the exam tomorrow and I don't want to be bogged down thinking about these limits if they're wrong. I got most of the problems correctly and it's only these two that's been weird for me, I've tried what I can but I'm just stumped. I'm currently googling ways to rewrite the GIF function.

$\endgroup$
  • $\begingroup$ Are you certain that, for the first one, it is not $x\to 2^+$? (For the second one, for any $x\in (1,\sqrt{2})$, the numerator is $0$ and the denominator is $x^2-1>0$, so the quantity is $0$. So the overall limit will be $0$, since the function is identically zero on $(1,1+\epsilon)$ for some $\epsilon > 0$). $\endgroup$ – Clement C. Aug 24 '16 at 13:29
  • $\begingroup$ Yep, our module only has answers for odd-numbered questions and the question before that one has $x \to 2^+$ which I know is $-\infty$ that's why I dismissed the answer key as wrong for the first question until I saw the next limit that gave me problems. $\endgroup$ – R. Cruz Aug 24 '16 at 13:31
  • $\begingroup$ For the second one, think of it this way: $\lim_{x\to1^+} \frac{0}{x^2-1} = ?$ $\endgroup$ – Clement C. Aug 24 '16 at 13:32
  • $\begingroup$ I see, shouldn't I care that the limit of denominator is 0 and thus the entire function has the form $\frac{0}{0}$? edit: I think I see what you're saying, it's not $\frac{0}{0}$ exactly but $\frac{0}{0^+}$? $\endgroup$ – R. Cruz Aug 24 '16 at 13:34
  • $\begingroup$ The limit of the denominator is zero, but that is not relevant since the numerator is exactly zero. Even if you multiply $0$ by a quantity increasingly big, the product will still be exactly $0$ (not that you don't "multiply by $\infty$": you multiply by something which tends to $\infty$) $\endgroup$ – Clement C. Aug 24 '16 at 13:35
1
$\begingroup$

For the first one, you're right and the answer key would be for $x \to 2^+$. To see why, observe that we can work in the interval $(1, 2)$. Here, $\lfloor x \rfloor - 1 = 0$, and so the limit is $0$ if the denominator is not $0$. Indeed, that's exactly what happens, since $x - \lfloor x \rfloor$ is the fractional part function, that's $0$ only at the integers. Since there are no integers in $(1, 2)$, the limit is $0$.

For the second limit, we can apply the same reasoning with the interval $(1, \sqrt 2)$. Here, the numerator is identically zero, while the denominator is never zero. Hence, the limit evaluates to $0$ once again.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.