Question

$\underset{x\to 0}{lim}\frac{\left\{\sin \right(\alpha +\beta )x+\sin (\alpha -\beta )x+\sin 2\alpha x\}}{{\cos }^2\beta x-{\cos }^2\alpha x}$.

Answer 1

$\sin (\alpha +\beta )x+\sin (\alpha -\beta )x=2\sin \left(\frac{2\alpha x}{2}\right)\cos \left(\frac{2\beta x}{2}\right)$

$=2\sin \left(\alpha x\right)\cos \left(\beta x\right)$

and $\sin 2\alpha x=2\sin \alpha x\cos \alpha x$

and ${\cos }^2\beta x-{\cos }^2\alpha x=\frac{(\cos \beta x+\cos \alpha x)(\cos \beta x-\cos \alpha x)}{1+1}$ as $x\to 0$

$=2(-2\sin \left(\frac{\beta -\alpha }{2}\right)x\sin \left(\frac{\beta +\alpha }{2}\right)x)$

$\Rightarrow L=\underset{x\to 0}{lim}\frac{2\sin \alpha x}{-4\sin \left(\frac{\beta -\alpha }{2}\right)x\sin \left(\frac{\beta +\alpha }{2}\right)x}$

As $\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1$

$\Rightarrow \underset{\theta \to 0}{lim}\sin \theta =\theta$

$\Rightarrow L=\underset{x\to 0}{lim}\frac{2\alpha x}{-4\left(\frac{\beta -\alpha }{2}\right)\left(\frac{\beta +\alpha }{2}\right)x^2}=\frac{2\alpha }{(\alpha -\beta )(\alpha +\beta )x}=$ Not defined.