if a bounded sequence xn in [a,b] converges to x0 then x0 belongs to [a,b] If a bounded sequence xn in [a,b] converges to x0 then prove that x0 belongs to [a,b]. I don't understand here how can x0 belong to [a,b], it may be slightly greater than b too. Please help.
The sequence is convergent and thus bounded, so there exists $M$ and $g$ such that

$$
|a_{n}|&lt;M\qquad\text{and}\qquad\lim_{n\rightarrow\infty}a_{n}=g.
$$

This holds because we can find $M$ such that $[a,b]\in[-M,M].$ Then 

$$
|g|=|\lim_{n\rightarrow\infty}a_{n}|=\lim_{n\rightarrow\infty}|a_{n}|&lt;\lim_{n\rightarrow\infty}M=M
$$
So
$$
g\in[-M,M]\qquad\text{and}\qquad{a_{n}}\in[-M,M],
$$
as desired.
A convergent sequence is surely bounded. Since $a\le x_n\le b$, permanence of sign tells you that
$$
a\le\lim_{n\to\infty}x_n\le b
$$

If you want a detailed proof, assume by contradiction that $x_0&lt;a$. Then there exists $N$ such that, for $n&gt;N$,
$$
|x_n-x_0|&lt;\frac{a-x_0}{2}
$$
which implies
$$
x_n&lt;x_0+\frac{a-x_0}{2}=\frac{a+x_0}{2}&lt;a
$$
Contradiction.

Similarly for proving that $x_0\le b$.