Prove that as $ \lim_{x \to a} f(x) = L, \,\text{then}\, \lim_{x \to a} |f(x)| =|L|$

Question

Basically so far I have managed to break it down to the following :

$|f(x)-l|< \varepsilon$

then

$||f(x)|-|L||\leq|f(x)-L|< \varepsilon$

then squeeze

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

I'm not too confident with this type of proofs, so a push in the right direction would be much appreciated.

Answer 1

Using the limit definition: we have $\lim\limits_{x\to a}f(x)=L$ so for $\epsilon>0$ we find $\delta>0$ such that $|x-a|<\delta$ imply $|f(x)-L|<\epsilon$ but with your inequality which we can deduced it from the triangle inequality we have $||f(x)|-|L||\le|f(x)-L| <\epsilon$ hence we have $$\lim\limits_{x\to a}|f(x)|=|L|$$

Answer 2

You can use $\varepsilon-\delta$ definition:

$$ \lim_{x \to a} f(x) = L \implies |f(x)-L|< \varepsilon \,\text{whenever} \,|x-a| < \delta$$ and so $$ ||f(x)|-|L|| < |f(x)-L| < \varepsilon \,\,\text{whenever} \,|x-a| < \delta \implies \lim_{x \to a} |f(x)| = |L|$$

Answer 3

You have done it right. The proof is completed.