Proof verification of $\lim_{x \to x_0} ( f(x) g(x) ) = km $ if $\lim_{x \to x_0}f(x) = k $ and $ \lim_{x \to x_0} g(x) = m$

Question

Suppose $f: A \subset \mathbb{R} \to \mathbb{R} $ and $g: A \subset \mathbb{R} \to \mathbb{R} $ such that $f(x) \to k$ and $g(x) \to m$. Let $\epsilon > 0$ for some $\epsilon \in \mathbb{R} $. Since $f(x) \to k$ and $g(x) \to m$, there are $\delta_f$ and $\delta_g$ such that $$0 < |x - x_0| <\delta_f \Rightarrow |f(x) - k| < \epsilon $$ and $$0 < |x - x_0| <\delta_f \Rightarrow |g(x) - m| < \epsilon $$. Let $\delta = min\{\delta_f, \delta_g\} $. Then we have: $$ 0 < |x - x_0| <\delta \Rightarrow |f(x) - k||g(x) - m| < \epsilon^2 $$. Observe now that, for any fixed $\epsilon$, both $f$ and $g$ are limited in this $\epsilon$-neighborhood of $x_0$. Therefore there are $L_f$ and $L_g$ (which depend on $\epsilon$ too) such that $\forall x (|f(x)| < L_f) \land \forall x (|g(x) < L_g)$, with $x \in B_\delta (x)\backslash \{x_0\} \subset A$.

We now expand the product $|f(x) - k||g(x) - m|$ (I will denote $f(x)$ by $f$, the same for $g(x)$, for the sake of my tired fingers): $$(1): \; |fg -kg -mf +km|;$$ Let $ C = (L_g k + L_f m)$. Then: $$(2): \; |fg - C/2 - C/2 +km| < \epsilon + |C| $$ $$(3): \; |fg - C/2 - (C/2 -km)| < \:...$$ Since $|a - b| ≥ |a|-|b|$, $$(4): \; |fg - C/2| - |C/2 -km| <\: ...$$ $$(5): \; |fg| - |C/2| - (|C/2| - |km|) <\:...$$ $$(6): \; |fg|+|km| < \epsilon + 2|C|$$ Since $|x| = |-x|$, $$(7): \; |fg| + |-km| < \epsilon + 2|C|$$ Since $|x + y| ≤ |x| + |y|$, $$(8): \; |fg - km| < \epsilon + 2|C|$$ Since each step respect the fact that the left side of the inequality is always decreasing, we conclude by $(8)$ that $fg \to km$.

Answer 1

Assume wlog $k,m> 0$, then we have that

• $|f(x)-k|<\epsilon_1 \iff 0<k-\epsilon_1 \le f(x) \le k+\epsilon_1$
• $|g(x)-m|<\epsilon_2\iff 0<m-\epsilon_2 \le g(x) \le m+\epsilon_2$

then

$$ (k-\epsilon_1)(m-\epsilon_2) \le f(x)g(x) \le (k+\epsilon_1)(m+\epsilon_2)$$

$$ km-k\epsilon_2-m\epsilon_1+\epsilon_1\epsilon_2 \le f(x)g(x) \le km+k\epsilon_2+m\epsilon_1+\epsilon_1\epsilon_2 $$

that is

$$|f(x)g(x)-km|\le \epsilon_3$$

with $\epsilon_3=|k\epsilon_2+m\epsilon_1-\epsilon_1\epsilon_2|$.

Answer 2

For any $\epsilon>0$ there exist $\delta_f,\delta_g>0$ such that

$$0 < |x - x_0| <\delta_f \Rightarrow |f(x) - k| < \epsilon $$

$$0 < |x - x_0| <\delta_g \Rightarrow |g(x) - m| < \epsilon$$

So, if we consider $\delta=\min\{\delta_f,\delta_g\}$ we have

$$0 < |x - x_0| <\delta \Rightarrow |f(x) - k| < \epsilon $$

$$0 < |x - x_0| <\delta \Rightarrow |g(x) - m| < \epsilon$$

Now

\begin{align}|f(x)g(x)-km| &=|f(x)g(x)-kg(x)+kg(x)-km|\\ &\le |f(x)g(x)-kg(x)|+|kg(x)-km|.\end{align} Thus if $0 < |x - x_0| <\delta$ then $$|f(x)g(x)-km|\le |g(x)||f(x)-k|+|k||g(x)-m|<\epsilon(|k|+|g(x)|). $$

Since

$$0 < |x - x_0| <\delta \Rightarrow |g(x) - m| < \epsilon\implies |g(x)|<|m|+\epsilon$$ we have

$$0 < |x - x_0| <\delta \implies |f(x)g(x)-km|<\epsilon (|k|+|m|+\epsilon). $$ If we consider $\epsilon<1$ it is

$$0 < |x - x_0| <\delta \implies |f(x)g(x)-km|<\epsilon (|k|+|m|+1)$$ and we are done.