Question

решите пожалуйста срочно

Answer 1

$limlimits _{x to 0}frac{arcsin^25x}{1-cos5x}=limlimits _{x to 0}frac{(5x)^2}{2sin^2frac{5x}{2}}=limlimits _{x to 0}frac{25x^2}{2cdot (frac{5x}{2})^2}=limlimits _{x to 0}frac{25x^2}{frac{25x^2}{2}}=2\\\star ; ; arcsinalpha (x)sim alpha (x); ; ,; ; sinalpha (x)sim alpha (x); ; ,; ; alpha (x)to 0; ; star$