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I want to prove that $\lim_{x \to a} f(x)g(x) = \lim_{x \to a} f(x) \lim_{x \to a} g(x)$. Is this a valid/good/correct proof?

Let $\lim_{x \to a} f(x) = L_1$ and $\lim_{x \to a} g(x) = L_2$. The righthand side can be written as $L_1L_2$.

Then $(\lim_{x \to a} f(x) - L_1) = \lim_{x \to a} f(x) - \lim_{x \to a} L_1 = L_1 - L_1 = 0$ and $(\lim_{x \to a} g(x) - L_2) = \lim_{x \to a} g(x) - \lim_{x \to a} L_2 = L_2 - L_2 = 0$.

Let $\epsilon > 0$. There exists a $\delta_1$ such that $|f(x)-L_1| < \sqrt{\epsilon}$ whenever $0 < |x-a| < \delta_1$, and there exists a $\delta_2$ such that $|g(x)-L_2| < \sqrt{\epsilon}$ whenever $0 < |x-a| < \delta_2$.

Therefore, suppose $0 < |x-a| < \delta$ where $\delta = \min(\delta_1, \delta_2)$ such that $0 < |x-a| < \delta_1$ and $0 < |x-a| < \delta_2$ are both satisfied. Then $|(f(x)-L_1)(g(x)-L_2)| < \sqrt{\epsilon}\sqrt{\epsilon} = \epsilon$.

Consider the expansion $(f(x)-L_1)(g(x)-L_2) = f(x)g(x) - L_2f(x) - L_1g(x) + L_1L_2$. Rearranged, we get $f(x)g(x) = (f(x)-L_1)(g(x)-L_2) + L_2f(x) + L_1g(x) - L_1L_2$.

We now have:

$\lim_{x \to a} f(x)g(x) = \lim_{x \to a}((f(x)-L_1)(g(x)-L_2) + L_2f(x) + L_1g(x) - L_1L_2)$

$\lim_{x \to a} f(x)g(x) = \lim_{x \to a}(f(x)-L_1)(g(x)-L_2) + \lim_{x \to a} L_2f(x) + \lim_{x \to a}L_1g(x) - \lim_{x \to a}L_1L_2$

$\lim_{x \to a} f(x)g(x) = 0 + L_1L_2 + L_1L_2 - L_1L_2$

$\lim_{x \to a} f(x)g(x) = L_1L_2$

Masacroso
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user525966
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1 Answers1

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You simplify $\lim_{x \to a}(f(x)-L_1)(g(x)-L_2)$ to 0, but isn't that using the fact that $f(x)-L_1$ and $g(x)-L_2$ go to zero, thus their product goes to zero, and therefore relies on the very thing you're trying to prove? You also use lim(f(x)+g(x)) = lim(f(x))+lim(g(x)) repeatedly. Have you proved that claim?

In this sort of question, you're being asked to analyze a limit at low, epsilon-delta level, so you should be focusing n that level rather than working a higher, "this limit simplifies to this" level. A general tactic with limit is to define error terms equal to the value minus the limit: error1 = f(x)-L1, error2 = f(x)-L2. Then

f(x)g(x) = (L1-error1)(L2-error2) = L1*L2-L1*error2-L2*error1+error1*error2

and

f(x)g(x)-L1*L2 = -L1*error2-L2*error1+error1*error2

Now you just have to show that this expression can be made arbitrarily small.

Acccumulation
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  • "You also use lim(f(x)+g(x)) = lim(f(x))+lim(g(x)) repeatedly. Have you proved that claim?" in a separate proof yes – user525966 Feb 28 '18 at 19:36
  • "A general tactic with limit is to define error terms equal to the value minus the limit" isn't this what i am doing though? – user525966 Feb 28 '18 at 19:37