Question

$\lim _{\alpha \rightarrow \beta}\left[\frac{\sin ^2 \alpha-\sin ^2 \beta}{\alpha^2-\beta^2}\right]$ is equal to

(a) 0

(b) 1

(c) $\frac{\sin \beta}{\beta}$

(d) $\frac{\sin 2 \beta}{2 \beta}$

Answer 1

$\lim _{\alpha \rightarrow \beta} \frac{\sin ^2 \alpha-\sin ^2 \beta}{\alpha^2-\beta^2}$, applying L'Hospital's rule

$\begin{array}{l}=\lim _{\alpha \rightarrow \beta} \frac{2 \sin \alpha \cos \alpha-0}{2 \alpha-0} \\=\lim _{\alpha \rightarrow \beta} \frac{\sin 2 \alpha}{2 \alpha} \\=\frac{\sin 2 \beta}{2 \beta}\end{array}$

So, The correct option of this question will be (D).