Ta biết rằng $\csc x=\frac{1}{\sin x }\Rightarrow(\csc x)' = -\frac{\cos x}{\sin^ 2x}$. Do đó theo quy tắc Lopitan
$\mathop {\lim }\limits_{x \to 0+}\frac{\csc x}{\ln x}=\mathop {\lim }\limits_{x \to 0+}\frac{(\csc x)' }{(\ln x)'}=\mathop {\lim }\limits_{x \to 0+}\frac{-\frac{\cos x}{\sin^ 2x}}{\frac{1}{x}}=\mathop {\lim }\limits_{x \to 0+}-\frac{x\cos x}{\sin^ 2x}$
$=\mathop {\lim }\limits_{x \to 0+}-\left (\frac{x}{\sin x}\right )^2.\cos x .\frac{1}{x}=-1.1.(+\infty)=-\infty$.