Given:
[tex]\lim_{x\to0}\text{ }\frac{sin^2x\text{ + sinx\lparen cosx -1\rparen}}{x^2}[/tex]
Apply L' Hospital's rule:
[tex]=\lim_{x\to0}\text{ }\frac{sin(2x)\text{ + cosx\lparen cosx - 1\rparen-sin}^2x}{2x}[/tex]
Apply L'Hospital's rule:
[tex]=\lim_{x\to0}\text{ }\frac{2cos(2x)\text{ -2sin\lparen2x\rparen+ sin\lparen x\rparen}}{2}[/tex]
Plug the value of 0 for x:
[tex]\begin{gathered} =\text{ }\frac{2cos0\text{ - 2sin0 + sin0}}{2} \\ =\text{ }\frac{2}{2} \\ =\text{ 1} \end{gathered}[/tex]
Answer: 1
