Question

$\underset{\alpha \to \beta }{lim}\frac{{\sin }^2\alpha -{\sin }^2\beta }{{\alpha }^2-{\beta }^2}$

• A. $0$
• B. $1$
• C. $\frac{\sin \beta }{\beta }$
• D. $\frac{\sin 2\beta }{2\beta }$

Answer 1

The correct option is D $\frac{\sin 2\beta }{2\beta }$

$si{n}^2\alpha -si{n}^2\beta =sin(\alpha +\beta ).sin(\alpha -\beta )$

Also ${\alpha }^2-{\beta }^2=(\alpha +\beta )(\alpha -\beta )$

As ${lim}_{\Theta \to 0}\frac{sin}{\theta }=1\Rightarrow {lim}_{\alpha \to \beta }\frac{sin(\alpha -\beta )}{\alpha -\beta }=1$

${lim}_{\alpha \to \beta }\frac{si{n}^2\alpha -si{n}^2\beta }{{\alpha }^2{\beta }^2}{lim}_{\alpha \to \beta }=\frac{sin(\alpha +\beta )}{\alpha +\beta }=\frac{sin\left(2\beta \right)}{2\beta }\Rightarrow \left(D\right)$