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I apologize if this is a stupid question, but I don't really understand this part of limits. I know that for all functions $f(x)$ and $g(x)$ and all numbers $a$,

$$\lim_{x \to a} f(x)g(x) = \lim_{x \to a} f(x) \lim_{x \to a} g(x).$$

I'm not sure if we can substitute other limits into the limits for the individual function, though. For example, say

$$\lim_{x \to a} f(x) = 0$$

and

$$\lim_{x \to a} g(x) = \infty.$$

The limit is obviously of indeterminate form. But what if we consider the limit of a constant multiple of $f(x)$? Wouldn't that be $0$ also?

$$\lim_{x \to a} cf(x) = \lim_{x \to a} f(x)$$

But in that case, can't we substitute the first limit in for the second in the limit of the products so that

$$\lim_{x \to a} f(x)g(x) = \lim_{x \to a} cf(x)\lim_{x \to a} g(x) = c\lim_{x \to a} f(x)\lim_{x \to a} g(x).$$

What am I doing wrong here? Thank you!

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Your first statement is false. If $\lim\limits_{x\to a}f(x)$ and $\lim\limits_{x\to a}g(x)$ exist, then the limit of the product is the product of the limits. This seems to be what you're missing. You can only 'substitute limits in' if they actually exist individually.

Take for example $f(x) = x$ and $g(x) = \dfrac{a}{x}$ where $a\neq 0$. While $\lim\limits_{x\to 0}f(x)g(x) = a$ (in particular, it exists), $\lim\limits_{x\to 0}g(x)$ does not exist.

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  • $\begingroup$ Wow, I should have realized that! I am really confused today. Thanks! $\endgroup$ – anon Sep 9 '13 at 22:46
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Note that your first equation is true if $\lim_{x\to a} f(x)=A$ and $\lim_{x\to a} g(x)=B$ with $A,B \in \mathbb{R}$, i.e. these two limits exist.

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  • $\begingroup$ I should have noticed that; thank you so much. $\endgroup$ – anon Sep 9 '13 at 22:46

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