Then $\lim_{z\rightarrow 0} f(z)$ choose the correct option?

Question

Let $f : \mathbb{C^∗} \rightarrow \mathbb{C}$ be the function defined by $f(z) := \ z \sin( \frac{1}{z})$. Then $\lim_{z\rightarrow 0} f(z)$

choos the correct option

$(a)$ is equal to zero.

$(b)$ does not exist.

$(c)$ is infinite.

$(d)$ is finite but not equal to zero.

My attempts : i thinks the option $d)$ will be correct

$\lim_{z\rightarrow 0} f(z)=\lim_{z\rightarrow 0}\ z \sin( \frac{1}{z})=1$

Is my answer correct or not ?

any hints/solution will be appreciated

Answer 1

If $z$ is complex.

Suppose $z$ were completly imaginary

$z = iy$ (with $y$ real)

$f(iy) = iy\sin \frac {1}{iy} = -iy\sin i\frac {1}{y} = y\sinh \frac {1}{y}$

$\lim_\limits{y\to 0} f(iy) = \infty$

If $z$ is real.

$\lim_\limits{x\to 0} f(x) = 0$

The limit does not exist.

Alternatively the Laurent series for $z\sin \frac{1}{z} = 1 - \frac {1}{6} z^{-2} + \frac {1}{5!} z^{-4} + \cdots$

And that limit does not exist as $z$ approaches $0.$

Answer 2

If $t$ is positive real number, and if $z=t$, then $f(z)=t\sin\ \frac{1}{t}$ goes to $0$

If $z=it$, then $f(z)= it \frac{e^{1/t} - e^{-1/t} }{2i} $. Note that $$ \lim_{t>0,\ t\rightarrow 0}\ \frac{te^{1/t}}{2} =\infty$$ so that limit does not exist.