Let $(x_n)$ be a sequence of real numbers.
I am wondering we can formally prove that:
$$ \lim_{n\rightarrow\infty}|x_n| < a$$
for all $a > 0$, implies
$$ \lim_{n\rightarrow\infty}|x_n| = 0$$
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Sign up to join this communityLet $(x_n)$ be a sequence of real numbers.
I am wondering we can formally prove that:
$$ \lim_{n\rightarrow\infty}|x_n| < a$$
for all $a > 0$, implies
$$ \lim_{n\rightarrow\infty}|x_n| = 0$$
Let $L=\lim_{n\rightarrow\infty}|x_n|$. Then $0\le L<a $ for all $a>0$. Suppose $L>0$. Then $\frac{L}{2}>0$ and so using $a=\frac{L}{2}$ gives $$L<a=\frac{L}{2}\Rightarrow L<0$$ which is a contradiction
Suppose $\varepsilon > 0$ is the limit under consideration. Choosing $a=\frac{\varepsilon}{2}$ gives us a contradiction from your first equation. Therefore $\varepsilon$ has to be smaller or equal to zero.
Now, from your first equation, and because there is a $\delta$ such that for $n> n_0$ all the elements of $|x_n|$ belong to $]0, \delta[$, we have to have the limit of $|x_n|$ belonging to $[0, \delta]$, i.e. it has to be non-negative.
The only non negative number which is smaller or equal to zero is zero.
q.e.d.