A Graphical Limit I was wondering what the answer is to this question. I'm sure that the answer is A or E and the answer given by our teacher is A. My logic is that 0 * undefined = defined?? but could anyone explain intuitively what/why the answer is, and what the graph may look like?
Thank you!
https://i.stack.imgur.com/ikwQb.png
 A: Your teacher is right.
In general, we do not always have that $\lim_{x \to a} F(x) G(x) = \left( \lim_{x \to a} F(x)\right)\left(\lim_{x \to a} G(x)\right)$. Specifically, this does not necessarily hold if either one of $F$ or $G$ does not have a limit (the limit is undefined).
So, we should use the definition of the limit instead. Consider a point $x = 1 + h$, with $h > 0$. We see that $f(1 + h)$ will be close to $0$ if $h$ is small. We also see that $g(2 + h)$ will be $-1$. So, the product of the two will be very close to $0$.
Likewise, if $h < 0$, we see that $f(1 + h)$ is very close to $0$ if $h$ is close to $0$, and $g(2+h) = 1$. So, their product is very close to $0$.
You can even formalize this further, using ($\epsilon$, $\delta$) definitions and such, but the above already makes it clear that the limit coming from either direction is equal, so the limit exists.
A: Your teacher's right and the answer is A
$g(x+1) = 1$ for $x\le 1$
$g(x+1) = -1$ for $x>1$
So $f(x)g(x+1)$ looks like $f(x)$ as x goes from negative infinity to 1.
At $x=1$, $f(x)g(x+1) = 2$
For $x>1$, $f(x)g(x+1)$ looks like f(x) but flipped above the x-axis.
So the graph goes down and then back up. But with a hole at (1,0). And a dot at (1,2).
From the graph it is clear that the limit of $f(x)g(x+1)$ as x goes to 1 exists and is 0.
