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Let $f(z)= [\frac {x^2} {x^2 + y^2}] + 2i $. Does f have a limit at $z=0$? [HINT: Investigate {$f(z_n)$} for sequences {$z_n$} approaching 0 along the real and imaginary axes separately.

SOLUTION:

No. Obeserve that although $ \frac 1 n \rightarrow 0$ and $\frac i n \rightarrow0$ as $n \rightarrow 0$, $f(\frac 1 n) \rightarrow 1 + 2i $ and $ f(\frac i n) \rightarrow 2i$; thus $lim_{x \rightarrow 0} f(z) $ does not exist.


I'm not quite sure how they are taking a limit of this function that seems to be not be admissible (written in terms of z). I see it as $ \frac {Re(z)^2} {|z|}$, which doesn't exist... but I don't see the logic behind their method of using sequences.

I'm not sure how they chose $\frac 1 n$ ,or $\frac i n$ or why they chose them. Overall, I'm not entirely sure what they are doing at all in terms of even writing {$f(z_n)$} and what that even is, or how they are even examining $\frac 1 n$ and $\frac i n$ as inputs for the function when there is no z present in the function they have written.

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2 Answers 2

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If a limit as $z \to 0$ exists, one should be able to plug in any sequence $\{ z_n \}$ going to zero and get the same limit. Limits of sequences are generally easier to work with. So in this case if the limit existed then $f(1/n)$ and $f(i/n)$ would go to the same thing, but they don't.

As to $x$, $y$ instead of $z$, that is surely because they use a convention that $z = x+iy$. Whichever book it is that you are using, they should mention that somewhere, possibly early on. It is a common convention. So $x = \text{Re}\, z$ and $y = \text{Im}\, z$ and you have the function written in terms of $z$.

You are correct that it is $\frac{(\text{Re}\, z)^2}{|z|^2} + 2i$, but how do you know that has no limit? You have to somehow show that too.

As to how do you choose such a sequence $1/n$ and $i/n$? There is no rule for this. In this case we are just looking at the function along the real and the imaginary axis. $1/n$ is a very common choice for a sequence of positive real numbers going to zero.

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  • $\begingroup$ Okay, so when we are evaluating $f(1/n)$ for example how would that look? Like $ lim_{n \rightarrow \infty} \frac {Re(\frac 1 n)^2} {|\frac 1 n|} + 2i $? How would you manage to know what $Re(\frac 1 n)$ is? I understand the rest of your answer, and I greatly appreciate it. $\endgroup$ Commented Sep 5, 2016 at 22:19
  • $\begingroup$ $(1/n)$ is real as $n$ is a natural number (that is, 1,2,3,4,5,...), so $Re(1/n) = 1/n$ for all natural numbers. On the other hand $i/n = i(1/n)$ so it is purely imaginary. When you see $1/n$ and taking the limit we're just looking at the value of the function at $1, 1/2, 1/3, 1/4, 1/5, 1/6, ...$ and seeing where it goes. $\endgroup$
    – Jiri Lebl
    Commented Sep 5, 2016 at 22:21
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If we let $z=x+iy$ tend to zero along this sequence:

$z_0=1+i$

$z_1=\frac{1}{10}+\frac{1}{100}i$

$z_2=\frac{1}{100}+\frac{1}{10000}i$

$z_3=\frac{1}{1000}+\frac{1}{1000000}i$

and in general $z_n=\frac{1}{10^n}+\frac{1}{100^n}i$

...then I think we would find that $f(z_n)$ tends to 1 since $y$ gets dwarfed by $x$.

On the other hand, if we let $z=x+iy$ tend to zero using the sequence $z_n=\frac{1}{100^n}+\frac{1}{10^n}i$ then I think we would find that $f(z_n)$ tends to zero since $x$ gets dwarfed by $y$.

From what I remember at uni, we only allow ourselves to say "$f(z)$ has a limit as $z$ tends to blah" if the limit is the same along all sequences ${z_n}$ that tend to blah.

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