MHB Considering the criterion for limit proofs

[Amad27]

In a proof.

Prove that **given**:

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

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

We know that

$$|f(x) - L| < \epsilon \space \text{for} \space |x - a| < \delta_1$$

What is the objective then?

Do we prove there exists a $\delta_2$ such that $\displaystyle \lim_{x\to a} |f(x)| = |L|$

Or do we go from the fact that**it is true that:** $|f(x) - L| < \epsilon \space \text{for} \space |x - a| < \delta_1$ and then somehow using $\delta_1$ derive that:

$| |f(x)| - |L| | < \epsilon$

Or do we find a $\delta$?We can easily state:

$$\left| |f(x) - |L| \right| \le \left| f(x) - L \right| < \epsilon$$

But we did not find the $\delta$?

Bottomline: So what is the point? Finding $\delta$ or just proving it is $< \epsilon$ assuming there will be a $\delta$ so that the statement will be true?

Thanks!

 

[Siron]

It's not necessary to work with another $\delta_2>0$. So yes, you just assume there will be a $\delta$ so that the statement will be true.

Proof
Let $\varepsilon>0$, since $\lim_{x \to a} f(x)=L$ there exists a $\delta>0$ such that $||f(x)|-|L||\leq |f(x)-L|<\epsilon$ if $0<|x-a|<\delta$.

Conclusion: we have found for every $\varepsilon>0$ a $\delta>0$ such that $||f(x)|-|L||<\varepsilon$ if $0<|x-a|<\delta$.