MHB Limits of Complex Functions .... Final Remark from Palka in Section 2.2 ....

[Math Amateur]

I am reading Bruce P. Palka's book: An Introduction to Complex Function Theory ...

I am focused on Chapter 2: The Rudiments of Plane Topology ...

I need help with some aspects regarding an example in Palka's final remarks in Section 2.2 Limits of Functions ...

Palka's final remarks in Section 2.2 which include the example read as follows:View attachment 7372I have two questions regarding the above text ...Question 1

In the above text from Palka Section 2.2 we read the following:" ... ... Consider for a moment, the function $$f(x) = e^{-1/x^2}$$. We are aware from calculus that $$\text{lim}_{x \rightarrow 0} \ f(x) = \text{lim}_{x \rightarrow 0} \ e^{-1/x^2} = 0$$. ... ..."Can someone please demonstrate exactly how/why $$\text{lim}_{x \rightarrow 0} \ f(x) = \text{lim}_{x \rightarrow 0} \ e^{-1/x^2} = 0$$ ... ... ?Question 2

In the above text from Palka Section 2.2 we read the following:

" ... ... However, if we set $$z_n = (2n \pi)^{-1/2} e^{ \pi i / 4 }$$ for $$n = 1,2, \ ... $$ , we observe that $$z_n \rightarrow 0$$, whereas $$f(z_n) = e^{ 2n \pi i } = 1$$ for every $$n$$ so $$f( z_n ) \rightarrow 1$$. ... ... "I am unable to show that $$f(z_n) = e^{ 2n \pi i }$$ ... can someone please help ...Help will be much appreciated ...

Peter

 

[Opalg]

Question 1

Can someone please demonstrate exactly how/why $$\text{lim}_{x \rightarrow 0} \ f(x) = \text{lim}_{x \rightarrow 0} \ e^{-1/x^2} = 0$$ ... ... ?

Starting from the fact that $e^t \to \infty$ as $t\to\infty$, it follows that $\lim_{t\to\infty}e^{-t} = \lim_{t\to\infty}\dfrac1{e^t} = \dfrac1{\lim_{t\to\infty}e^{t}} = 0.$ If you then put $t = 1/x^2$, noticing that $t\to\infty$ as $x\to0$, it follows that $\lim_{x\to0}e^{-1/x^2} = 0.$

Question 2

In the above text from Palka Section 2.2 we read the following:

" ... ... However, if we set $$z_n = (2n \pi)^{-1/2} e^{ \pi i / 4 }$$ for $$n = 1,2, \ ... $$ , we observe that $$z_n \rightarrow 0$$, whereas $$f(z_n) = e^{ 2n \pi i } = 1$$ for every $$n$$ so $$f( z_n ) \rightarrow 1$$. ... ... "I am unable to show that $$f(z_n) = e^{ 2n \pi i }$$ ... can someone please help ...

If $$z_n = (2n \pi)^{-1/2} e^{ \pi i / 4 }$$ then $$z_n^2 = (2n \pi)^{-1} e^{ \pi i / 2 }$$. But $e^{ \pi i / 2 } = i$. So $z_n^2 = \dfrac i{2n\pi}$, and $\dfrac1{z_n^2} = \dfrac{2n\pi}i = -2n\pi i.$ Finally, $$f(z_n) = e^{-1/z_n^2} = e^{ 2n \pi i } = 1.$$

 

[Math Amateur]

Starting from the fact that $e^t \to \infty$ as $t\to\infty$, it follows that $\lim_{t\to\infty}e^{-t} = \lim_{t\to\infty}\dfrac1{e^t} = \dfrac1{\lim_{t\to\infty}e^{t}} = 0.$ If you then put $t = 1/x^2$, noticing that $t\to\infty$ as $x\to0$, it follows that $\lim_{x\to0}e^{-1/x^2} = 0.$If $$z_n = (2n \pi)^{-1/2} e^{ \pi i / 4 }$$ then $$z_n^2 = (2n \pi)^{-1} e^{ \pi i / 2 }$$. But $e^{ \pi i / 2 } = i$. So $z_n^2 = \dfrac i{2n\pi}$, and $\dfrac1{z_n^2} = \dfrac{2n\pi}i = -2n\pi i.$ Finally, $$f(z_n) = e^{-1/z_n^2} = e^{ 2n \pi i } = 1.$$

Thanks for the help, Opalg ...

Peter