Question

${lim}_{x\to 0}\frac{\left\{sin\right(\alpha +\beta )x+sin(\alpha -\beta )x+sin2\alpha x\}}{co{s}^2\beta x-co{s}^2\alpha x}$

Answer 1

${lim}_{x\to 0}\frac{\left\{sin\right(\alpha +\beta )x+sin(\alpha -\beta )x+sin2\alpha x\}}{co{s}^2\beta x-co{s}^2\alpha x}$

$={lim}_{x\to 0}\frac{2sin\left[\frac{(\alpha +\beta )x+(\alpha -\beta )x}{2}\right]cos\left[\frac{(\alpha +\beta )x-(\alpha -\beta )x}{2}\right]+2sin\alpha xcos\alpha x}{(1-si{n}^2\beta x)-(1-si{n}^2\alpha x)}$

$={lim}_{x\to 0}\frac{2sin\alpha xcos\beta x+2sin\alpha xcos\alpha x}{si{n}^2\alpha x-si{n}^2\beta x}$

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

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

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

$=\frac{2\alpha \times 1\times (1+1)}{{\alpha }^2\times 1-{\beta }^2\times 1}\times {lim}_{x\to 0}\frac{1}{x}$

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