lim⁡θ→0(1−cos(2θ)sin2(5θ))\lim _{\theta \rightarrow 0}\left(\frac{1-\textrm{cos}\left(2\theta \right)}{\textrm{sin}^{2}\left(5\theta \right)}\right)

asked by guest
on Jan 27, 2025 at 5:18 am



You asked:

Evaluate the expression: lim⁡θ→01−cos⁡(2θ)sin⁡(5θ)2\lim_{θ \to 0}{\frac{1 - \cos\left( 2 θ \right)}{{\sin\left( 5 θ \right)}^{2}}}

MathBot Answer:

lim⁡θ→01−cos⁡(2θ)sin⁡(5θ)2=225\lim_{θ \to 0}{\frac{1 - \cos\left( 2 θ \right)}{{\sin\left( 5 θ \right)}^{2}}} = \frac{2}{25}
    • Axis Limits
    • min x
    • max x
    • min y
    • max y
  • Update Axis Limits
Close Controls

+ Submit Question