Question

The value of $\underset{\alpha \to \beta }{lim}\left[\frac{{\sin }^2\alpha -{\sin }^2\beta }{{\alpha }^2-{\beta }^2}\right]$ is:

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

Answer 1

Answer: (C)

$\underset{\alpha \to \beta }{lim}\left[\frac{{\sin }^2\alpha -{\sin }^2\beta }{{\alpha }^2-{\beta }^2}\right]=\underset{\alpha \to \beta }{lim}\left[\frac{\sin (\alpha -\beta )\sin (\alpha +\beta )}{(\alpha -\beta )(\alpha +\beta )}\right]$

$=\underset{\alpha \to \beta }{lim}\left[\frac{\sin (\alpha -\beta )}{(\alpha -\beta )}\right]\times \underset{\alpha \to \beta }{lim}\left[\frac{\sin (\alpha +\beta )}{(\alpha +\beta )}\right]$

$=\underset{\alpha -\beta \to 0}{lim}\left[\frac{\sin (\alpha -\beta )}{(\alpha -\beta )}\right].\left(\frac{\sin 2\beta }{2\beta }\right)$ ....... $[\because \alpha \to \beta ⟹\alpha -\beta \to 0]$

$=1.\frac{\sin 2\beta }{2\beta }$ ..... $[\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1]$

$=\frac{\sin 2\beta }{2\beta }$