Use the $\varepsilon$-$\delta$ definition of a limit to prove this.

Question

I know to how prove normal limits using the epsilon-delta definition, say:

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

But, there was a question on my textbook which I couldn't quite figure out to do, even though I've thought about it for a while I don't even know how to go about starting it.

Use $\varepsilon$-$\delta$ definition of a limit to prove that

$$\lim \limits _{x\to c}f(x) = 0$$ iff $$\lim \limits_{x\to c}|f(x)| = 0$$

Could anyone help me with this, even a hint on where to start? Thank you in advance.

Answer 1

With $L=0$ you wish to prove that $$\forall \varepsilon >0\exists\delta >0 \forall x\in D_f(|x-c|<\delta\implies |f(x)-L|<\varepsilon) \\ \Updownarrow \\ \forall \varepsilon >0\exists\delta >0 \forall x\in D_{|f|}(|x-c|<\delta\implies ||f(x)|-L|<\varepsilon)$$

which is equivalent to proving that $$\forall \varepsilon >0\exists\delta >0 \forall x\in D_f(|x-c|<\delta\implies |f(x)|<\varepsilon) \\ \Updownarrow \\ \forall \varepsilon >0\exists\delta >0 \forall x\in D_{|f|}(|x-c|<\delta\implies |\color{green}|f(x)\color{green}||<\varepsilon).$$

Now you need to prove that $D_f=D_{|f|}$, once this is done your initial problem becomes equivalent to proving $$\forall \varepsilon >0\exists\delta >0 \forall x\in D_f(|x-c|<\delta\implies |f(x)|<\varepsilon) \\ \Updownarrow \\ \forall \varepsilon >0\exists\delta >0 \forall x\in D_f(|x-c|<\delta\implies |\color{green}|f(x)\color{green}||<\varepsilon).$$

Now just prove that for all $x\in D_f$ the equality $|\color{green}|f(x)\color{green}||=|f(x)|$ holds and you get

$$\forall \varepsilon >0\exists\delta >0 \forall x\in D_f(|x-c|<\delta\implies |f(x)|<\varepsilon) \\ \Updownarrow \\ \forall \varepsilon >0\exists\delta >0 \forall x\in D_f(|x-c|<\delta\implies |f(x)|<\varepsilon)$$

which is obviously true.