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Question

Question: \[\lim_{\alpha \rightarrow \beta}\frac{\sin^{2}\alpha - \sin^{2}\beta}{\alpha^{2} - \beta^{2}}\]...

lim⁡α→βsin⁡2α−sin⁡2βα2−β2\lim_{\alpha \rightarrow \beta}\frac{\sin^{2}\alpha - \sin^{2}\beta}{\alpha^{2} - \beta^{2}}

A

0

B

1

C

sin⁡ββ\frac{\sin\beta}{\beta}

D

sin⁡2β2β\frac{\sin 2\beta}{2\beta}

Answer

sin⁡2β2β\frac{\sin 2\beta}{2\beta}

Explanation

Solution

lim⁡α→βsin⁡2α−sin⁡2βα2−β2\lim_{\alpha \rightarrow \beta}\frac{\sin^{2}\alpha - \sin^{2}\beta}{\alpha^{2} - \beta^{2}}=lim⁡α−β→0sin⁡(α+β)sin⁡(α−β)(α+β)(α−β)\lim_{\alpha - \beta \rightarrow 0}\frac{\sin(\alpha + \beta)\sin(\alpha - \beta)}{(\alpha + \beta)(\alpha - \beta)}

=lim⁡α−β→0sin⁡(α−β)(α−β)lim⁡α−β→0sin⁡(α+β)(α+β)\lim_{\alpha - \beta \rightarrow 0}\frac{\sin(\alpha - \beta)}{(\alpha - \beta)}\lim_{\alpha - \beta \rightarrow 0}\frac{\sin(\alpha + \beta)}{(\alpha + \beta)} =lim⁡α→βsin⁡(α+β)(α+β)\lim_{\alpha \rightarrow \beta}\frac{\sin(\alpha + \beta)}{(\alpha + \beta)}

=sin⁡2β2β\frac{\sin 2\beta}{2\beta}.

Trick : By L’ Hospital’s rule, lim⁡α→β2sin⁡αcos⁡α2α=sin⁡2β2β\lim_{\alpha \rightarrow \beta}\frac{2\sin\alpha\cos\alpha}{2\alpha} = \frac{\sin 2\beta}{2\beta}.