Does the limit exist ? (AP Calculus)

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 $h(x) g(x) = 0$.

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 $0$.

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.