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$$\lim_{x \rightarrow a}f(x)=L_1 , \lim_{x\rightarrow a}g(x) = L_2 $$

I'm trying to prove

$$\lim_{x \to a} f(x) g(x) = L_1 L_2$$

We have that

$$|f(x)g(x)-L_1L_2| = |f(x)g(x)-g(x)L_1+g(x)L_1 -L_1 L_2 |$$

Factoring $g(x)$ and $L_1$ and using triangle identity $$|g(x)(f(x)-L_1)+L_1(g(x)-L_2)| \leq |g(x)\color{blue}{(f(x)-L_1)}| + |L_1\color{red}{(g(x)-L_2)}|$$

Let $|f(x)-L_1|<\epsilon_1$ and $|g(x)-L_2| < \epsilon_2$ then we can have that

$$\epsilon_1 = \dfrac{\epsilon}{|g(x)|}, \epsilon_2 = \dfrac{\epsilon}{L_1}$$

However, I've been told that I cannot divide epsilon by a non-constant value like $g(x)$. Does it look right?

kobe
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Melz
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1 Answers1

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No, the argument is incorrect. Given $\epsilon > 0$, you need to find a corresponding $\delta > 0$ such that for all $x$, $0 < |x - a| < \delta$ implies $|f(x)g(x) - L_1L_2| < \epsilon$. Without loss of generality, assume $0 < \epsilon < 1$. Choose $\delta_1 > 0$ such that $|f(x) - L_1| < \epsilon/(1 + |L_1| + |L_2|)$ whenever $0 < |x - a| < \delta_1$, and choose $\delta_2 > 0$ such that $|g(x) - L_2| < \epsilon/(1 + |L_1| + |L_2|)$ whenever $0 < |x - a| < \delta_2$. Set $\delta = \min\{\delta_1,\delta_2\}$ and show that $|f(x)g(x) - L_1L_2| < \epsilon$ whenever $0 < |x - a| < \delta$. You will need to use the fact that $|g(x)| < 1 + |L_2|$ for $0 < |x - a| < \delta$, which follows from the triangle inequality.

kobe
  • 41,901