MAT334-2018F > Quiz-2

Q2 TUT 5301

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Victor Ivrii:
Find the limits as $z\to \infty$ of the given function, or explain why it
does not exist:
\begin{align*}
\end{align*}

Junya Zhang:
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.

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

hanyu Qi:
In attachment.

Victor Ivrii:
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
And no uncommon abbreviations like DNE.