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limit x tends to 0 (1/x^2 - 1/(sinx)^2 )

limit x tends to 0 (1/x^2 - 1/(sinx)^2 )

Grade:11

1 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

Dear tushar

Ltx→0(1/x^2 - 1/(sinx)^2 )

or

Ltx→0(sin2x-x2)/x2sin2x

or Ltx→0(sin2x-x2)/x4{sin2x/x2}

or Ltx→0(sin2x-x2)/x4     because  Ltx→0 sinx/x     =1

Use L hosipal rule

 limit will be =-1/3


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