# limit $\sqrt{-n^4+4n^2+4}-in^2 = -2i$?

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!

• You did $\infty-\infty=0$ at the end. – logarithm May 13 '19 at 3:15
• I think your mistake is more or less like a complex version of saying that $\infty-\infty=0$ – Julian Mejia May 13 '19 at 3:15
• If you multiply and divide by $\sqrt{-n^4+4n^2+4}+in^2$, then it becomes $\frac{4n^2+4}{\sqrt{-n^4+4n^2+4}+in^2}=\frac{4+4/n^2}{\sqrt{-1+4/n^2+4/n^2}+i}\to\frac{4}{2i}=-2i$. – logarithm May 13 '19 at 3:20
• 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 $\infty$. – trancelocation May 13 '19 at 5:46

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 $$n\to\infty$$.

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

$$n^4-4n^2-4= (n^2-2)^2-8$$;

$$\sqrt{(n^2-2)^2-8} - (n^2-2)-2$$;
Set $$m:=n^2-2$$, and consider
$$\lim_{ m \rightarrow \infty} (\sqrt{m^2-8}-m -2)$$;