Question

Lim ln(1-sinx)/arcsin5x
x-->0

Answer 1

По правилу Лопиталя
$lim_{x to0} frac{ln(1-sinx)}{arcsin5x}={frac{0}{0}}=lim_{x to0} frac{(ln(1-sinx))'}{(arcsin5x)'}= lim_{x to0} frac{ frac{-cosx}{1-sinx}}{ frac{5}{ sqrt{1-(5x)^2} }}=$
$=lim_{x to0} frac{-cosx sqrt{1-(5x)^2}}{ 5(1-sinx)}=frac{-cos0 sqrt{1-(5*0)^2}}{ 5(1-sin0)}=-frac{1}{ 5};$