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.
Best Answer
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.