Singularity of $f(z)=\frac{\sin z}{z}$ at $z=0$ I'm reading Conway's complex analysis book and on page 110 he asked to determine the nature of the singularity at $z=0$ of the function $f(z)=\frac{\sin z}{z}$ and if it's a removable singularity he asked to define $f(0)$ so that $f$ is analytic at $z=0$.
Question 1:
If this function has an isolated singularity at $z=0$ I know how to prove this one is removable (it suffices to prove $\lim_{z\to 0}zf(z)=0$ from theorem 1.2 of the same section). The problem is why does this function have an isolated singularity at $z=0$ in the first place?
Question 2:
I have a solution manual that says the function is analytic at $z=0$ if we define $f(0)=1$, because $\lim_{z\to 0}\frac{\sin z}{z}=1$ (why?)
 A: Clearly the function 
$$
f(z)=\frac{\sin(z)}{z}
$$
is holomorphic on $\mathbb C \backslash \{0\}$. Therefore $f$ has an isolated singularity at $z=0$ which is removable since
$$
\lim_{z\to 0}zf(z)=\lim_{z\to 0}z\frac{\sin(z)}{z}=\lim_{z\to 0}\sin(z)=0
$$
So we can remove this singularity. Now we just need to extend $f$ in $z=0$ continuously which we can do using a special case of the l'Hopital rule (check here for example), so:
$$
\lim_{z\to 0}f(z)=\lim_{z\to 0}\frac{\sin(z)}{z}=\lim_{z\to 0}\cos(z)=1
$$
Therefore by setting $f(0)=1$ we have continuously extended $f$ on $\mathbb C$ and removed the singularity.
A: It has an isolated singularity at $0$ because dividing by $0$ is undefined and thus $f(z)$ is not defined at $0$.
Second, there are two ways to think about this.
1) Since we already accept that this singularity is removable, the only way to define $f$ at $0$ is by its limit for else it would not be continuous and thus we would not have removed the singularity.
2) Going along the lines of the proof, $g(z)=zf(z)=sin(z)=z-\frac{z^3}{3!}+\ldots$.
Then, $g(z)=z(1-\frac{z^2}{3!}+\ldots)=zh(z)$ where $h(z)=1-\frac{z^2}{3!}+\ldots$ is analytic. 
Thus, as in the proof, we have to define $f(0)=h(0)=1$.
