Epsilon Delta limit of absolute values?

Question

I'm pretty new to Epsilon Delta limits, and I came across a problem that asked me to take the absolute value of a function. Namely $lim_{x\to a} \mid f(x) - g(x)\mid$, where $lim_{x\to a} f(x) = C, \lim_{x\to a} g(x) = D$. How should I go about proving that the limit is $\mid C-D \mid$, using the Epsilon Delta definition of a limit? Thanks!

Answer 1

We need to show that $$\forall\varepsilon>0, \exists\delta>0\text{ such that }0<|x-a|<\delta\implies||f(x)-g(x)|-|C-D||<\varepsilon.$$

We know that

\begin{align*} &\forall\varepsilon>0,\exists\delta_1>0\text{ such that }0<|x-a|<\delta_1\implies|f(x)-C|<\frac\varepsilon2\text{, and}\\ &\forall\varepsilon>0,\exists\delta_2>0\text{ such that }0<|x-a|<\delta_2\implies|g(x)-D|<\frac\varepsilon2 \end{align*}

Question 1: Why are we allowed to use $\frac\varepsilon2$ instead of $\varepsilon$?

Set $\delta=\min\{\delta_1,\delta_2\}$, and suppose $0<|x-a|<\delta$. Then the above two statements still hold with $\delta$ instead of $\delta_1$ or $\delta_2$.

Question 2: Why do the two above statements hold for $\delta$?

Now,

\begin{align*} ||f(x)-g(x)|-|C-D||&\leq|(f(x)-g(x))-(C-D)|\\ &=|(f(x)-C)-(g(x)-D)|\\ &\leq|f(x)-C|+|g(x)-D|\\ &<\frac\varepsilon2+\frac\varepsilon2\\ &=\varepsilon. \end{align*}

Question 3: Why is $||f(x)-g(x)|-|C-D||\leq|(f(x)-g(x))-(C-D)|$?