lim z tends to i (z^2+1/z-i)
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Step-by-step explanation:
To start with, note that limn→∞(1+an)bn=eab . This will almost certainly come up again and again, so you will need to memorise this.
In this question, a=−2 and b=1 , therefore limx→∞(1−2n)n=e−2
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