$(a_n)_{n=1}^\infty$ is a convergent sequence and $a_n \in [0,1]$ for all $n$. Proof of limit $(a_n)_{n=1}^\infty$ lies in [0,1]. Textbook question:

$(a_n)_{n=1}^\infty$ is a convergent sequence and $a_n \in [0,1]$ for all $n$. Proof of limit $(a_n)_{n=1}^\infty$ lies in [0,1].

I don't understand the question I suppose. It would seem that it answers itself...? If $a_n \in [0,1]$ for all $n$, then of course the limit does since we know it is convergent...?
What am I missing?

Assume that the limit of $a_n$ don't live in $[0,1]$. Without loss of generality assume that limit $L$ is $L<0$ (i.e. is at the left of the interval). Convergence says $\forall \epsilon>0 \exists N, \forall n>N |a_n-L|<\epsilon $. Take $\epsilon=|\frac{L}{2}|=-\frac{L}{2}$ then in the interval $\left(\frac{3L}{2},\frac{L}{2}\right)$ (which is not in the interval $[0,1]$) we must find all the elements of $a_n$ from $n\geq N$, then is a contradiction because all the elements of $a_n$ lives in $[0,1]$ 
