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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.

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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.

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    $\begingroup$ What you are misunderstanding about limits is that you can not simply split apart a product inside of the limits as a product outside of limits. $\lim\limits_{x\to c}(a(x)\times b(x))$ is not the same thing as $\lim\limits_{x\to c}(a(x))\times \lim\limits_{x\to c}(b(x))$ and this is a perfect example of that. $\endgroup$ – JMoravitz Sep 15 '17 at 19:37
  • $\begingroup$ Even though $f$ fails to be continuous at $1$, it's limit exists and is $0$. $\endgroup$ – Mark Viola Sep 15 '17 at 19:38
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    $\begingroup$ 1) Yes $\lim\limits_{x\to 1}f(x)g(x+1)$ exists and it is equal to $0$. 2) No this doesn't imply that $\lim\limits_{x\to 1}g(x)$ exists. 3) Yes $\lim\limits_{x\to 2}h(x)=\lim\limits_{x\to 1}h(x+1)$ for any function $h$ where the limit exists and in particular this is true for $g$ as well if the limit were to have existed. Since the limit doesn't exist however, it doesn't really make sense to use an equals sign here. $\endgroup$ – JMoravitz Sep 15 '17 at 19:41
  • $\begingroup$ @JMoravitz But the limit of the product does equal the product of of the limits.. at least it does if the limits of both functions exist and are finite. If not, then you might have more work to do. $\endgroup$ – Doug M Sep 15 '17 at 19:46
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    $\begingroup$ You got the part $0\times\text{ DNE }=\text{DNE} $ wrong $\endgroup$ – Paramanand Singh Sep 16 '17 at 2:46
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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.

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  • $\begingroup$ I like that you solved it without using squeeze theorem. $\endgroup$ – Michael Maliszesky Sep 16 '17 at 20:13
  • $\begingroup$ Yes, did i have to? $\endgroup$ – neonpokharkar Sep 16 '17 at 20:18
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    $\begingroup$ After thinking about this. I decided I like this way to solve the best. I know the squeeze theorem is the most popular approach. However I think this approach is easier for high school students to understand. And I always like solutions the use only the fundamental definitions. THANKS!! $\endgroup$ – Michael Maliszesky Sep 17 '17 at 17:56
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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 $h(x) g(x) = 0$.

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  • $\begingroup$ I see, i realize this is squeeze theorem. $\endgroup$ – Michael Maliszesky Sep 16 '17 at 20:06
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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 $0$.

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  • $\begingroup$ Ok, now that I think about it saying, "The limit of a product is the product of the limits" only make sense when everything is defined $\endgroup$ – Michael Maliszesky Sep 16 '17 at 20:10
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$\lim |f(x)g(x + 1)| = \lim |f(x)||g(x + 1)| = \lim|f(x)| = 0$, 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 $0$.

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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.

enter image description here

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.

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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.

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