I was so lost on this limits question. help please

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I just had a midterm for my calc class and there was this one question that I spent a good 20 minutes on and could not figure out. can someone help me with the solution please?
lim x→0 (sin x²)/sin²x

I can do it using L'Hopitals rule, but that wasn't allowed :rolleyes:
 
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lim x→0 (sin x²)/sin²x = lim x→0 [(sin x²)/x²][x²/sin²x]
 
Use the Maclaurin's series for sinx and cosx.
 

Thread 'Finding the nth roots of a complex number'

There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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