Question

Помогите решить предел

Answer 1

$limlimits_{x to 0}frac{sin2x}{x^2} ={frac{0}{0}}=|sin2x sim 2x|= limlimits_{x to 0}frac{2x}{x^2}=limlimits_{x to 0}frac{2}{x}=frac{2}{0} =infty$