Consider the interval $[0,1]$. Each element of it is a limit point, i.e. $\alpha$ is a limit of the sequence $n_1=\alpha, n_2=\alpha, \ldots$. Only $0,1$ are boundary points.

Well, as someone has figured it out by supplying the definitions of limit point and boundary point. Now if we just head toward the general set topological approach we will find that , if $\Bbb{S}$ ${\subset}$ of $\Bbb{R}$ , and if $\Bbb{X}$ be the boundary then $\Bbb{X}$=cl(S)~int ( S) . So if p is a boundary point, then p will be in $\Bbb{X}$ . And we call $\Bbb{S}$ a closed set if it contains all it's boundary points. Now as we also know it's equivalent definition that s will be a closed set if it contains all it limit point.

But that doesn't not imply that a limit point is a boundary point as a limit point can also be a interior point . Let's check the proof.

Let $\Bbb{S}$ is our set of which l is a int point . Then for `$\epsilon$>0 , N(l, $\epsilon$ ) contained in l . Now we will try to prove it contrapositively . Let l is not an int point . Then N(l, $\epsilon$ ) is not contained in $\Bbb{S}$ . Now let, €>0 then either €< $\epsilon$ or €≥ $\epsilon$

When, €< $\epsilon$ as N(l, $\epsilon$ ) is not contained in $\Bbb{S}$ , so N(l, €) is not also contained in s . It suggests that, N'(l,€) ${\cap}$ $\Bbb{S}$ = $\phi$ So, l is not a limit point of $\Bbb{S}$

When, €≥ $\epsilon$ , N(l, $\epsilon$ ) is contained in N(l, €). So from here also it can be shown that , $\Bbb{S}$ ${\cap}$ N'(l,€) is $\phi$ . So l is not a limit point of $\Bbb{S}$ .

So if l is not an int point of $\Bbb{S}$ , it's not an limit point of $\Bbb{S}$ . It implies that if l is an limit point of $\Bbb{S}$ , it's an interior point of $\Bbb{S}$ .

Now, there are also some cases where the above assertion fails. So l may or may not belongs to cl ( $\Bbb{S}$ ) ~ int ( $\Bbb{S}$ )

And the whole discussion tells us that a limit point can be a boundary point but that doesn't mean every limit point is a boundary point. And that's it !!!