Question

$\underset{x\to 0}{lim}\frac{x\cos x-\sin x}{x^2\sin x}$.

Answer 1

$\underset{x\to 0}{Lim}$ $\frac{xcosx-sinx}{x^{2}sinx}$

Applying L'Hospital's rule :

$\underset{x\to 0}{Lim}$ $\frac{-xsinx-cosx+cosx}{2xsinx+x^{2}cosx}$

$=\underset{x\to 0}{Lim}$ $\frac{-sinx-xcosx}{2sinx+2xcosx+2xcosx-x^{2}sinx}$

$=\underset{x\to 0}{Lim}$ $\frac{-cosx-cosx+xsinx}{2cosx+2cosx-2xsinx+2cosx+2xsinx-2xsinx-x^{2}cosx}$

$=\frac{-1-1}{2+2+2}=\frac{-2}{6}=\frac{-1}{3}$