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What is the limit of the following complex sequence? I get a different result than Wolfram Alpha. My approach was:

$$\sqrt{-n^4+4n^2+4}-in^2 = \sqrt{i^2n^4+4n^2+4}-in^2 = in^2\sqrt{1-\frac{1}{4n^2}-\frac{4}{n^4}}-in^2 \xrightarrow{n\xrightarrow{}\infty} 0$$

Wolfram Alpha says the limit is $-2i$. I'm confused. I must've made a stupid mistake, but cannot find it. Thanks in advance!

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    $\begingroup$ You did $\infty-\infty=0$ at the end. $\endgroup$ – logarithm May 13 '19 at 3:15
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    $\begingroup$ I think your mistake is more or less like a complex version of saying that $\infty-\infty=0$ $\endgroup$ – Julian Mejia May 13 '19 at 3:15
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    $\begingroup$ If you multiply and divide by $\sqrt{-n^4+4n^2+4}+in^2$, then it becomes $\frac{4n^2+4}{\sqrt{-n^4+4n^2+4}+in^2}=\frac{4+4/n^2}{\sqrt{-1+4/n^2+4/n^2}+i}\to\frac{4}{2i}=-2i$. $\endgroup$ – logarithm May 13 '19 at 3:20
  • $\begingroup$ The limit depends on whether you choose $\sqrt{-1} = i$ or $\sqrt{-1} = -i$. You will get either $-2i$ as shown by logarithm in the comment or you will get complex $\infty$. $\endgroup$ – trancelocation May 13 '19 at 5:46
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Let's forget about those pesky $i$s and consider the limit of $$n^2\sqrt{1-\frac4{n^2}-\frac4{n^4}}-n^2$$ as $n\to\infty$. This equals $$\frac{n^4\left(1-\frac4{n^2}-\frac4{n^4}\right)-n^4}{n^2\sqrt{1-\frac4{n^2}-\frac4{n^4}}+n^2} =\frac{-4n^2-4}{n^2\sqrt{1-\frac4{n^2}-\frac4{n^4}}+n^2} =\frac{-4-4/n^2}{\sqrt{1-\frac4{n^2}-\frac4{n^4}}+1} $$ which tends to $-2$ as $n\to\infty$.

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    $\begingroup$ (+1) I wrote a very similar answer... too similar. $\endgroup$ – robjohn May 13 '19 at 3:24
  • $\begingroup$ What about mentioning that the limit depends on the branch of $\sqrt{z}$? Taking $\sqrt{-1}=-i$ would lead to a different result. $\endgroup$ – trancelocation May 13 '19 at 4:14
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$n^4-4n^2-4= (n^2-2)^2-8$;

The expression reads:

$\sqrt{(n^2-2)^2-8} - (n^2-2)-2$;

Set $m:=n^2-2$, and consider

$\lim_{ m \rightarrow \infty} (\sqrt{m^2-8}-m -2)$;

Hence?

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  • $\begingroup$ In the last line it should me m^2 under root. $\endgroup$ – Milan May 13 '19 at 10:04
  • $\begingroup$ Milan.Yes!!Thanks, Greetings. $\endgroup$ – Peter Szilas May 13 '19 at 10:11

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