Evaluating trigonometric limit \(\displaystyle{{\csc}^{{2}}{\left({2}{x}\right)}}-{\frac{{{1}}}{{{4}{x}^{{2}}}}}\)

Kendall Daniels 2022-04-02 Answered
Evaluating trigonometric limit csc2(2x)14x2
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

arrebusnaavbz
Answered 2022-04-03 Author has 18 answers
Based on DeepSea's hint, I've managed to solve by myself.
limx0[csc2(2x)14x2]
=limx0[1sin2(2x)14x2]
=limx0[4x2sin2(2x)4x2(sin2(2x))]
=limx0[4x2sin2(2x)4x2×4x2(sin(2x)2x)2]
=limx0[4x2sin2(2x)4x2×4x2]×limx0[1(sin(2x)2x)2]
=limx0[4x2sin2(2x)16x4]×1
=limx0[8x2sin(4x)64x3]
=limx0[88cos(4x)192x2]
=limx0[1cos(4x)24x2]
=124×limx0[1cos(4x)x2]
=124×12×42
=13
Not exactly what you’re looking for?
Ask My Question
Lana Hamilton
Answered 2022-04-04 Author has 12 answers
hint: replace the denominator on the third line of your proof sin2(2x)  by  (2x)2 and apply L'hopitale rule three times to the expression: 4x2sin2(2x)16x4.I did it and it works. Try it. Note that "replace" here means you write: sin2(2x)=4x2(sin(2x)2x)2 and the second factor approaches 1 when x0
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

New questions