Proving that $x_n\to L$ implies $|x_n|\to |L|$, and what about the converse?

Solution 1:

For (ii) it's not necessary to split into $3$ cases: just write the definition using this inequality $$\left||x_n|-|L|\right|\le|x_n-L|$$ and for a counterexample take $$x_n=(-1)^n$$

Solution 2:

You know $|x_n-L|<\varepsilon$ no matter the sign of $L$.

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Proving that $x_n\to L$ implies $|x_n|\to |L|$, and what about the converse?

Solution 1:

Solution 2:

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