Krdytkyu

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 $fg to km$.

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

• $|f(x)-k|<epsilon_1 iff 0<k-epsilon_1 le f(x) le k+epsilon_1$


• $|g(x)-m|<epsilon_2iff 0<m-epsilon_2 le g(x) le m+epsilon_2$

then

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

$$ km-kepsilon_2-mepsilon_1+epsilon_1epsilon_2 le f(x)g(x) le km+kepsilon_2+mepsilon_1+epsilon_1epsilon_2 $$

that is

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

with $epsilon_3=|kepsilon_2+mepsilon_1-epsilon_1epsilon_2|$.

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

$$0 < |x - x_0| <delta Rightarrow |g(x) - m| < epsilonimplies |g(x)|<|m|+epsilon$$ 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.