Does the limit exist ? (AP Calculus) Below is a question from an AP Calculus exam. The answer key say choice C is the correct answer, so that implies that $$\lim_{x\to1} (f(x)g(x+1))$$ does exist. It seems to me that all the choices are true, and there is no correct answer. 

Question 1) If $\lim_{x\to1} (f(x)g(x+1))$ does exist then what is its value?
Question 2) Since it does exist, does that imply that $\lim_{x\to1} g(x+1)$ also exist?
Question 3) Isn't it true that:
$$\lim_{x\to1} g(x+1) = \lim_{x\to2} g(x)  $$ 
and it is established that $\lim_{x\to2} g(x) $ doesn't exist in choice (b)?

This is my reasoning:
$$\lim_{x\to1} (f(x)g(x+1))$$
$$[\lim_{x\to1}f(x)]  \times [\lim_{x\to1} g(x+1)]$$
$$[\lim_{x\to1}f(x)]   \times [\lim_{x\to2} g(x) ]$$
$$[0]  \times [DNE]$$
$$DNE$$
So there is either something I don't understand about limits, or the question is wrong. I want to say the question is wrong, but I'm not 100% confident. 
Please Help.
 A: The property that
$$ \lim_{x \to 1} f(x) g(x+1) = \lim f(x) \lim g(x+1) $$
is generically only true if both limits on the right exist. It is not always true.
In this case, it's clear that $g(x+1)$ is $1$ from the left and $-1$ from the right. So $f(x)g(x+1) = f(x)$ for $x < 1$ and $f(x)g(x+1) = -f(x)$ for $x > 1$. As $x \to 1$ (from either side), $f(x) \to 0$ and $-f(x) \to 0$, so the limit exists and is equal to $0$.
A: Based on the graphs, we can see for 
$$\lim_{x\to 1^+} f(x)g(x+1)$$$$=\lim_{x\to 1^+} f(x)\lim_{x\to 1^+} g(x+1)$$$$=0\times-1=0$$
$$$$$$$$$$\lim_{x\to 1^-} f(x)g(x+1)$$$$=\lim_{x\to 1^-} f(x)\lim_{x\to 1^-} g(x+1)$$$$=0\times 1=0$$
So the limit exists.
A: $\lim |f(x)g(x + 1)| = \lim |f(x)||g(x + 1)| = \lim|f(x)| = 0$, since $|g(x + 1)| = 1$ on a neighbourhood of $2$. Then, since the limit of the absolute value is 0, the original limit must be $0$.
A: The other answers are right, but I think that looking at the graph of $f(x)·g(x+1)$ helps to understand why the limit exists and it is zero.

Of course, this is a particular choice of f(x), but in fact any function with $\lim_{x\to 1}{f(x)}=0$ could work (as the other answers have proved).
Here $f(x)$ is in green, $g(x)$ in blue and $f(x)·g(x+1)$ in red. When we get close to 1, multiplying $f(x)$ by $g(x)$ just changes the sign of $f(x)$, but it keeps approaching 0 by both sides.
A: Note that $|g(x)| \leq 1$
$$0 \leq |f(x)g(x+1) | \leq |f(x)|$$
Now we can apply squeeze theorem and show that 
$$0 \leq \lim_{x \to 1} |f(x)g(x+1)| \leq \lim_{x \to 1} |f(x)| = 0 $$
We do not require $\lim_{x \to 1} g(x+1) $ to exists.
An extreme example would be $h(x) =0$ and $g(x)$ is some bounded function.  Regardless of what is $g(x)$ exactly, we always have $h(x) g(x) = 0$.
A: The limit exists because $f$ goes to zero and this compensates the discontinuity of $g$.
By the way, $g$ turns $f$ to $|f|$, and the absolute value preserves continuity.
