The associated legendre function is defined by
$$p_{\textit{l}}^{m\left ( x \right )}\equiv \left ( 1-x^{2} \right )^{|m|/2}\left ( \frac{\mathrm{d} }{\mathrm{d} x} \right )^{|m|}p_{\textit{l}}\left ( x \right )$$ where $$p_{\textit{l}}\left ( x \right )$$ is the $l$th legendre polynomial.
Now, for $$|m|>\textit{l},$$ we have $$p_{\textit{l}}^{m\left ( x \right )}=0$$
However, what I am unable to see is why for any given $l$, there are $$\left ( 2\textit{l}+1 \right )$$ possible values of $m$.
Could someone please explain?
No, you only have $l+1$ solutions from $$ p_{\textit{l}}^{m}\left ( x \right )=(-1)^m \left ( 1-x^{2} \right )^{m/2}\left ( \frac{\mathrm{d} }{\mathrm{d} x} \right )^{m}p_{\textit{l}}\left ( x \right )$$ for $m=0,1,\cdots,l$. For negative order, you actually use the relation $$ p_l^{-m}(x)=(-1)^m\frac{(l-m)!}{(l+m)!}p_l^{m}(x), $$ for $l=1,2,\cdots,l$.