4
$\begingroup$

enter image description here

Problem 3. Show that for a sequence $(x_n)$ the following are true:
(i) $\lim x_n=0$ if and only if $\lim |x_n|=0$.
(ii) $\lim x_n=L$ implies $\lim |x_n|=|L|$. Is the converse true? Prove or give a counterexample.

(i) is already done, easy.

I'm halfway done with (ii), I split it into three cases:

$L = 0$, $L > 0$ and $L < 0$.

For $L = 0$ I just refer to part (i).

For $L > 0$, we know that $|x_n - L| < \epsilon$

if lim $|x_n|$ = $|L|$, then we must have

$||x_n| -|L|| < \epsilon$

but by the reverse triangle equality, $||x_n| - |L|| < |x_n - L|$ so clearly $||x_n|-|L|| < \epsilon$, thus

lim $x_n = L$ $\implies$ lim $|x_n| = |L|$ for $L > 0$

for some reason I'm confused as to part three, $L<0$. I'm having trouble seeing the difference between $L>0$ and $L<0$, and i'm starting to think that splitting it up like that isn't necessary at all.

$\endgroup$
7
$\begingroup$

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 $$x_n=(-1)^n$$

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.