Question

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

Answer 1

${lim}_{x\to 0}\frac{\sin (\alpha +\beta )x+\sin (\alpha -\beta )x+\sin 2\alpha x}{{\cos }^2\beta x-{\cos }^2\alpha x}$

$=\underset{x\to 0}{lim}\frac{\frac{\sin (\alpha +\beta )x}{(\alpha +\beta )}\times (\alpha +\beta )+\frac{\sin (\alpha -\beta )x}{(\alpha -\beta )}\times (\alpha -\beta )+\frac{\sin 2\alpha x}{2\alpha }\times 2\alpha }{\sin (\alpha +\beta )\sin (\alpha -\beta )}$

$=\underset{x\to 0}{lim}\frac{\frac{\sin (\alpha +\beta )x}{(\alpha +\beta )}\times (\alpha +\beta )+\frac{\sin (\alpha -\beta )x}{(\alpha -\beta )}\times (\alpha -\beta )+\frac{\sin 2\alpha x}{2\alpha }\times 2\alpha }{\frac{\sin (\alpha +\beta )x}{(\alpha +\beta )}\times (\alpha +\beta )\frac{\sin (\alpha -\beta )}{(\alpha -\beta )}\times (\alpha -\beta )}$

$=\frac{\alpha +\beta +\alpha -\beta +2\alpha }{(\alpha +\beta )(\alpha -\beta )}$

$=\frac{4\alpha }{{\alpha }^2-{\beta }^2}$