<s>[INST] 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!
 [/INST] 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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