Question

$\underset{x\to 0}{\mathrm{lim}}\frac{1-\mathrm{cos}2x+{\mathrm{tan}}^{2}x}{x\mathrm{sin}x}$

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Solution

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{1-\mathrm{cos}2x+{\mathrm{tan}}^{2}x}{x\mathrm{sin}x}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 0}{\mathrm{lim}}\left[\frac{2{\mathrm{sin}}^{2}x+{\mathrm{tan}}^{2}x}{x\mathrm{sin}x}\right]\left[\because 1-\mathrm{cos}2\mathrm{A}=2{\mathrm{sin}}^{2}\mathrm{A}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{numerator}&\mathrm{denominator}\mathrm{by}{x}^{2}:\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\underset{x\to 0}{\mathrm{lim}}\left[\frac{\frac{2{\mathrm{sin}}^{2}x}{{x}^{2}}+\frac{{\mathrm{tan}}^{2}x}{{x}^{2}}}{\left(\frac{\mathrm{sin}x}{x}\right)}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{2{\left(1\right)}^{2}+{\left(1\right)}^{2}}{1}\left[\because \underset{x\to 0}{\mathrm{lim}}\frac{{\mathrm{sin}}^{2}x}{{x}^{2}}=1,\underset{x\to 0}{\mathrm{lim}}\frac{{\mathrm{tan}}^{2}x}{{x}^{2}}=1\right]\phantom{\rule{0ex}{0ex}}=3$

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