Author Topic: Q2 TUT 5301  (Read 1066 times)

Victor Ivrii

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Q2 TUT 5301
« 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*}

Junya Zhang

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Re: Q2 TUT 5301
« Reply #1 on: October 06, 2018, 10:37:45 AM »
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.
« Last Edit: October 06, 2018, 10:49:17 AM by Junya Zhang »

Victor Ivrii

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Re: Q2 TUT 5301
« Reply #2 on: October 06, 2018, 05:11:13 PM »
you are discussing $(x,y)\to 0$ which is not the case.
« Last Edit: October 08, 2018, 09:48:47 PM by Victor Ivrii »

hanyu Qi

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Re: Q2 TUT 5301
« Reply #3 on: October 08, 2018, 09:10:52 PM »
In attachment.

Victor Ivrii

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Re: Q2 TUT 5301
« Reply #4 on: October 08, 2018, 09:53:40 PM »
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.

Jeffery Mcbride

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Re: Q2 TUT 5301
« Reply #5 on: October 09, 2018, 08:49:26 PM »
PDF Attachment contains solution
« Last Edit: October 09, 2018, 08:52:08 PM by Jeff Mcbride »

Victor Ivrii

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Re: Q2 TUT 5301
« Reply #6 on: October 09, 2018, 09:48:50 PM »
Actually, Junya  dkd everything except going along $te^{i\theta}$ with $t\to +\infty$