# find lim x tends to 0 sin 2x + 3x / 4x - sin 5x

$\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\mathrm{sin}2\mathrm{x}+3\mathrm{x}}{4\mathrm{x}-\mathrm{sin}5\mathrm{x}}\phantom{\rule{0ex}{0ex}}=\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\frac{\mathrm{sin}2\mathrm{x}}{2\mathrm{x}}+\frac{3\mathrm{x}}{2\mathrm{x}}}{\frac{4\mathrm{x}}{5\mathrm{x}}-\frac{\mathrm{sin}5\mathrm{x}}{5\mathrm{x}}}×\left(\frac{2\mathrm{x}}{5\mathrm{x}}\right)\phantom{\rule{0ex}{0ex}}=\frac{2}{5}\frac{\underset{\mathrm{x}\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}2\mathrm{x}}{2\mathrm{x}}+\frac{3}{2}\right)}{\underset{\mathrm{x}\to 0}{\mathrm{lim}}\left(\frac{4}{5}-\frac{\mathrm{sin}5\mathrm{x}}{5\mathrm{x}}\right)}\phantom{\rule{0ex}{0ex}}=\frac{2}{5}×\frac{\left(1+\frac{3}{2}\right)}{\left(\frac{4}{5}-1\right)}\phantom{\rule{0ex}{0ex}}=\frac{2}{5}×\frac{-1}{2}\phantom{\rule{0ex}{0ex}}=\frac{-1}{5}$

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