Find lim x→0 (sin x²)/sin²x without l'Hôpital's rule

[vbplaya]

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'hospital's rule, but that wasn't allowed

 

[AKG]

lim x→0 (sin x²)/sin²x = lim x→0 [(sin x²)/x²][x²/sin²x]

 

[Fermat]

Use the Maclaurin's series for sinx and cosx.