Let $a_n$ be a sequence such that $$ \inf{a_n} = 2 $$ and $$ \lim{a_n} = 5 $$
Does exist $n_0$ such that $a_{n_0} = 2$?
I'm not sure how to approach this question, I believe that $2$ is in $a_n$, since there only exists one partial limit, and so if $2$ wasn't in $a_n$, then $\inf{a_n} > 2$ (since $a_n$ must be increasing?).
I'm not sure how I could justify myself, or if my argument is correct.
On
By definition $$\forall \epsilon>0 \mbox{, } \exists N \mbox{ s.t. } \forall n\ge N \mbox{, }|a_n-5|<\epsilon$$ so given $\epsilon=1$, $$\exists N \mbox{ s.t. } \forall n\ge N \mbox{, }|a_n-5|<1$$ so for all of these $n$, $4<a_{n}$ so there exist only other $N-1$ possible terms that are under $4$.
This means that $\inf a_n = \inf_{n<N}a_n=\min_{n<n_0}a_n=2 $ becouse of the finiteness of the set on which we are taking the $\inf$, so there exists by definition an element $a_{n_0}$ with $n_0<N$ such that $a_{n_0}=2$.
Hint: for large enough $n$ we have $a_n>3$. Consider the finitely many elements of the sequence before that point.