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?)
Best Answer
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.