+9657 \(\displaystyle\lim_{x\rightarrow 0}\dfrac{\sin ^2x - x}{\cos x - 1}\\ =\displaystyle\lim_{x\rightarrow 0}\dfrac{\sin 2x - 1}{-\sin x}\\ =-\dfrac{1}{0}\\ =\infty\)
Am I correct?
\(\mbox{Footnote:}\\ \dfrac{\mathtt d}{\mathtt dx}\sin^2x\\ u=\sin x\,\,\, y= u^2\\ \dfrac{\mathtt dy}{\mathtt dx}=\dfrac{\mathtt dy}{\mathtt du}\times \dfrac{\mathtt du}{\mathtt dx}\\ = 2u\times \cos x\\ = 2\sin x \cos x\\ = \sin 2x\)
