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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits2 Marks
MCQ
$\lim _{\alpha \rightarrow \beta} \frac{\sin ^2 \alpha-\sin ^2 \beta}{\alpha^2-\beta^2}=$
  • 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)
Applying L-Hospital's rule, we get
$\lim _{\alpha \rightarrow \beta} \frac{\sin ^2 \alpha-\sin ^2 \beta}{\alpha^2-\beta^2}=\lim _{\alpha \rightarrow \beta} \frac{2 \sin \alpha \cos \alpha}{2 \alpha}$
$=\lim _{\alpha \rightarrow \beta} \frac{\sin 2 \alpha}{2 \alpha}$
$=\frac{\sin 2 \beta}{2 \beta}$

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