Lim ln(1-sinx)/arcsin5xx-- больше 0

По правилу Лопиталя [latex]\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} }}=[/latex] [latex]=\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}; [/latex]