Question

Evaluate the following limits:

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

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\sin \left(\alpha +\beta \right)x+\sin \left(\alpha -\beta \right)x+\sin 2\alpha x}{{\cos }^2\beta x-{\cos }^2\alpha x} \\ =\underset{x\to 0}{\lim }\frac{2\sin \left[{\displaystyle \frac{\left(\alpha +\beta \right)x+\left(\alpha -\beta \right)x}{2}}\right]\cos \left[{\displaystyle \frac{\left(\alpha +\beta \right)x-\left(\alpha -\beta \right)x}{2}}\right]+2\sin \alpha x\cos \alpha x}{\left(1-{\sin }^2\beta x\right)-\left(1-{\sin }^2\alpha x\right)} \\ =\underset{x\to 0}{\lim }\frac{2\sin \alpha x\cos \beta x+2\sin \alpha x\cos \alpha x}{{\sin }^2\alpha x-{\sin }^2\beta x} \\ =\underset{x\to 0}{\lim }\frac{2\sin \alpha x\left(\cos \beta x+\cos \alpha x\right)}{{\sin }^2\alpha x-{\sin }^2\beta x} \end{gathered}$$
$$\begin{gathered} =\underset{x\to 0}{\lim }\frac{2\alpha x\times {\displaystyle \frac{\sin \alpha x}{\alpha x}}\times \left(\cos \beta x+\cos \alpha x\right)}{{\alpha }^2{x}^2\times {\displaystyle \frac{{\sin }^2\alpha x}{{\alpha }^2{x}^2}}-{\beta }^2{x}^2\times {\displaystyle \frac{{\sin }^2\beta x}{{\beta }^2{x}^2}}} \\ =\frac{2\alpha \times \underset{x\to 0}{\lim }{\displaystyle \left(\frac{\sin \alpha x}{\alpha x}\right)}\times \underset{x\to 0}{\lim }\left(\cos \beta x+\cos \alpha x\right)}{{\alpha }^2\times {\left(\underset{x\to 0}{\lim }{\displaystyle \frac{\sin \alpha x}{\alpha x}}\right)}^2-{\beta }^2\times {\displaystyle {\left(\underset{x\to 0}{\lim }\frac{\sin \beta x}{\beta x}\right)}^2}}\times \underset{x\to 0}{\lim }\frac{x}{x^2} \\ =\frac{2\alpha \times 1\times \left(1+1\right)}{{\alpha }^2\times 1-{\beta }^2\times 1}\times \underset{x\to 0}{\lim }\frac{1}{x} \\ =\frac{4\alpha }{{\alpha }^2-{\beta }^2}\times \infty \\ =\infty \end{gathered}$$

Disclaimer: There is some mistake or some misprinting in the question.