I was so lost on this limits question. help please

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    Limits Lost
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SUMMARY

The limit problem presented involves evaluating lim x→0 (sin x²)/sin²x without using L'Hôpital's rule. The solution can be approached by applying the Maclaurin series for sin(x) and cos(x). Specifically, the limit can be rewritten as lim x→0 [(sin x²)/x²][x²/sin²x], which simplifies the evaluation process. This method provides a clear pathway to the solution without relying on calculus techniques that are restricted in the context of the problem.

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  • Understanding of limits in calculus
  • Familiarity with the Maclaurin series expansion
  • Knowledge of trigonometric functions and their properties
  • Basic skills in algebraic manipulation of limits
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  • Explore alternative limit evaluation techniques beyond L'Hôpital's rule
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vbplaya
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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'hospital's 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.
 

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