Toronto Math Forum
MAT3342018F => MAT334Tests => Quiz2 => Topic started by: Victor Ivrii on October 05, 2018, 06:16:22 PM

Find the limits as $z\to \infty$ of the given function, or explain why it
does not exist:
\begin{align*}
&h(z)= \frac{z}{z},\qquad z\ne 0.
\end{align*}

The limit does not exist.
By definition of limit as $z\to\infty$,
$$\lim_{z\to\infty} h(z) =\lim_{z\to\infty} \frac{z}{z} = \lim_{z\to 0} \frac{\frac{1}{z}}{\frac{1}{z}} =\lim_{z\to 0} \frac{\frac{1}{z}}{\frac{1}{z}} = \lim_{z\to 0} \frac{z}{z} $$
Let $z = x + iy$, then
$$\lim_{z\to\infty} h(z) = \lim_{(x,y)\to (0,0)} \frac{x+iy}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to (0,0)} \frac{x}{\sqrt{x^2+y^2}} + i\frac{y}{\sqrt{x^2+y^2}} $$
Note that $ \lim_{(x,y)\to (0,0)} \frac{x}{\sqrt{x^2+y^2}} $ does not exist since $$ \lim_{(x,y)\to (0,0)} \frac{x}{\sqrt{x^2+y^2}} = 1 $$ when $z$ approaches 0 alone the positive real axis, and $$ \lim_{(x,y)\to (0,0)} \frac{x}{\sqrt{x^2+y^2}} = 1 $$ when $z$ approaches 0 alone the negative real axis.
Similarly, $ \lim_{(x,y)\to (0,0)} \frac{y}{\sqrt{x^2+y^2}} $ does not exist.
This implies that $\lim_{z\to\infty} h(z)$ does not exist.

you are discussing $(x,y)\to 0$ which is not the case.

In attachment.

Nice colour. I will delete it tomorrow since you have not scanned properly: to black and white.
http://forum.math.toronto.edu/index.php?topic=1078.0 (http://forum.math.toronto.edu/index.php?topic=1078.0)
And no uncommon abbreviations like DNE.

PDF Attachment contains solution

Actually, Junya dkd everything except going along $te^{i\theta}$ with $t\to +\infty$