Question

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

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Solution

## $\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(\alpha +\beta \right)x+\mathrm{sin}\left(\alpha -\beta \right)x+\mathrm{sin}2\alpha x}{{\mathrm{cos}}^{2}\beta x-{\mathrm{cos}}^{2}\alpha x}\phantom{\rule{0ex}{0ex}}=\underset{x\to 0}{\mathrm{lim}}\frac{2\mathrm{sin}\left[\frac{\left(\alpha +\beta \right)x+\left(\alpha -\beta \right)x}{2}\right]\mathrm{cos}\left[\frac{\left(\alpha +\beta \right)x-\left(\alpha -\beta \right)x}{2}\right]+2\mathrm{sin}\alpha x\mathrm{cos}\alpha x}{\left(1-{\mathrm{sin}}^{2}\beta x\right)-\left(1-{\mathrm{sin}}^{2}\alpha x\right)}\phantom{\rule{0ex}{0ex}}=\underset{x\to 0}{\mathrm{lim}}\frac{2\mathrm{sin}\alpha x\mathrm{cos}\beta x+2\mathrm{sin}\alpha x\mathrm{cos}\alpha x}{{\mathrm{sin}}^{2}\alpha x-{\mathrm{sin}}^{2}\beta x}\phantom{\rule{0ex}{0ex}}=\underset{x\to 0}{\mathrm{lim}}\frac{2\mathrm{sin}\alpha x\left(\mathrm{cos}\beta x+\mathrm{cos}\alpha x\right)}{{\mathrm{sin}}^{2}\alpha x-{\mathrm{sin}}^{2}\beta x}$ $=\underset{x\to 0}{\mathrm{lim}}\frac{2\alpha x×\frac{\mathrm{sin}\alpha x}{\alpha x}×\left(\mathrm{cos}\beta x+\mathrm{cos}\alpha x\right)}{{\alpha }^{2}{x}^{2}×\frac{{\mathrm{sin}}^{2}\alpha x}{{\alpha }^{2}{x}^{2}}-{\beta }^{2}{x}^{2}×\frac{{\mathrm{sin}}^{2}\beta x}{{\beta }^{2}{x}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{2\alpha ×\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\alpha x}{\alpha x}\right)×\underset{x\to 0}{\mathrm{lim}}\left(\mathrm{cos}\beta x+\mathrm{cos}\alpha x\right)}{{\alpha }^{2}×{\left(\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\alpha x}{\alpha x}\right)}^{2}-{\beta }^{2}×{\left(\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\beta x}{\beta x}\right)}^{2}}×\underset{x\to 0}{\mathrm{lim}}\frac{x}{{x}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2\alpha ×1×\left(1+1\right)}{{\alpha }^{2}×1-{\beta }^{2}×1}×\underset{x\to 0}{\mathrm{lim}}\frac{1}{x}\phantom{\rule{0ex}{0ex}}=\frac{4\alpha }{{\alpha }^{2}-{\beta }^{2}}×\infty \phantom{\rule{0ex}{0ex}}=\infty$ Disclaimer: There is some mistake or some misprinting in the question.

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