Can the limit of a sequence be $+\infty$? The proposition comes from the textbook of Tao Analysis:

Proposition 6.4.12. Let $(a_n)_{n=m}^∞ $ be a sequence of real numbers, let $L^+$
be the limit superior of this sequence, and let $L^-$ be the limit inferior of
this sequence (thus both $L^+$ and  $L^-$ are extended real numbers).
Let $c$ be a real number. If $(a_n)_{n=m}^∞ $  converges to $c$, then we must
have $L^+$ = $L^-$ = $c$. Conversely, if $L^+$ = $L^-$ = $c$, then $(a_n)_{n=m}^∞ $
converges to $c$.

My question is: Can we set c to be an extended real number? So if $L^+$ = $L^-$ = $c$=$+\infty$, then $(a_n)_{n=m}^∞ $
converges to $+\infty$. I'm just wondering if the limit of a sequence can be $+\infty$. If yes, then the sequence (1,2,3,$\dots$) has a limit of $+\infty$.
 A: Answering your questions in reverse order:

I'm just wondering if the limit of a sequence can be $+\infty$. If yes, then the sequence $(1,2,3,\ldots)$ has a limit of $+\infty$.

This is perfectly fine.  It's a true statement that $(1,2,3,\ldots)$ has a limit of $+\infty$ in the extended real numbers.

Can we set $c$ to be [any] extended real number? So if $L^+ = L^- = c=+\infty$, then $(a_n)_{n=m}^\infty$ converges to $+\infty$.

This, however, is not okay.  By convention, if a sequence of real numbers converges, then its limit is a real number.  Therefore, the phrase "$(a_n)_{n=m}^\infty$ converges to $+\infty$" is either false or nonsense.  In the former case, your statement following "So" is false when $L^-=c=+\infty$, and in the latter case, it's not even a logical statement.
You may write the following analogous statement for the infinite extended real case:

Let $c \in \{-\infty, +\infty\}$ be an extended real number. If $(a_n)_{n=m}^\infty$ diverges to $c$, then we must
have $L^+ = L^- = c$. Conversely, if $L^+ = L^- = c$, then $(a_n)_{n=m}^\infty$ diverges to $c$.

