Question

limit x=0,(1/x^2-1/sin^2x)

Answer 1

Using Puiseux series:

Lim x → 0 1\sinx²-1\x²=Lim x → 0 (1\x²+1/3+x²/15+O(x^4))-1\x²=

=Lim x → 0 1\x²-1\x²+1/3+x²/15+O(x^4)=Lim x → 0 0+1/3+0²/15+O(0^4)=

=Lim x → 0 1/3+0+0=1/3.

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