$\lim_{z\to -1} (z+1) \sin(\frac{1}{z+1})$, for complex variable $z$.

Question

I want to find this limit for complex variable $z$

$$\lim_{z\to -1} (z+1) \sin(\frac{1}{z+1})$$

In the real case I know $\sin(z)$ is bounded by $-1, 1,$ and the limit is $0$. But in the complex case $\sin(z)$ is not bounded. How can I find the limit or prove that it doesn't exist?

Answer 1

One may first consider the sequence $\left\{z_n \right\}$ of complex numbers such that $$ z_n+1=\frac1{n\pi}, \quad n \to \infty, \quad z_n \to -1, $$ giving $$ \lim_{z_n\to -1} (z_n+1) \sin\left(\frac{1}{z_n+1}\right)=\lim_{n\to \infty} \frac1{n\pi} \sin(n\pi)=0. $$ Then one may consider the sequence $\left\{z_n \right\}$ of complex numbers such that $$ z_n+1=\frac1{n\pi i}, \quad n \to \infty, \quad z_n \to -1, $$ giving $$ \lim_{z_n\to -1} (z_n+1) \sin\left(\frac{1}{z_n+1}\right)=\lim_{n\to \infty} \frac{e^{n\pi}-e^{-n\pi}}{2n\pi}=\infty. $$

Then considered limit $$\lim_{z\to -1} (z+1) \sin\left(\frac{1}{z+1}\right)$$ does not exist.