MCQ
$\mathop {\lim }\limits_{\alpha \to \beta } \left[ {\frac{{{{\sin }^2}\alpha - {{\sin }^2}\beta }}{{{\alpha ^2} - {\beta ^2}}}} \right] = $
  • A
    $0$
  • B
    $1$
  • C
    $\frac{{\sin \beta }}{\beta }$
  • $\frac{{\sin 2\beta }}{{2\beta }}$

Answer

Correct option: D.
$\frac{{\sin 2\beta }}{{2\beta }}$
d
(d) $\mathop {\lim }\limits_{\alpha \to \beta } \frac{{{{\sin }^2}\alpha - {{\sin }^2}\beta }}{{{\alpha ^2} - {\beta ^2}}}$

Applying $ L-$ Hospital’s rule,

$\mathop {{\rm{lim}}}\limits_{\alpha \to \beta } \frac{{2\sin \,\alpha \,\,\cos \alpha }}{{2\alpha }} = \mathop {{\rm{lim}}}\limits_{\alpha \to \beta } \frac{{\sin \,\,2\alpha }}{{2\alpha }} = \frac{{\sin \,\,2\beta }}{{2\beta }}$.

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