Question

Sol. [Key Concept: Expand $sin\ x = x - \frac{x^{3}}{6} + dots$; the numerator $sim\ x^{4}$ and the denominator $x^{2} - \sin^{2}xsim\frac{2x^{4}}{3!}$, so the limit is $3$.]

$$\lim_{x \rightarrow 0}\mspace{2mu}\frac{x^{2}\sin^{2}x}{x^{2} - \sin^{2}x} $$Applying expansion

$${\lim_{x \rightarrow 0}\mspace{2mu}\frac{x^{2}\sin^{2}x}{x^{2} - \left( x - \frac{x^{3}}{3!} + \ldots \right)^{2}} }{\lim_{x \rightarrow 0}\mspace{2mu}\frac{x^{2}\sin^{2}x}{x^{2} - x^{2} + \frac{2x^{4}}{3!}\ldots} = \frac{3!}{2} = 3 }$$