0
$\begingroup$

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:

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

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

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

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

$\endgroup$
  • $\begingroup$ as the answer below said you are using the statement to be proved in your proof, thus it is not correct. $\endgroup$ – Masacroso Feb 28 '18 at 17:57
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ "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 $\endgroup$ – user525966 Feb 28 '18 at 19:36
  • $\begingroup$ "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? $\endgroup$ – user525966 Feb 28 '18 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.