\frac{\pi}{6}) \frac{2sinx + 1}{4cos2x - 1} = \\ = \frac{2sin( \frac{\pi}{6} )+ 1}{4cos( \frac{\pi}{3} ) - 1} = \frac{ 2\times 0.5 + 1}{4 \times 0.5 - 1} = \\ = \frac{2}{1} = 2" alt="lim(x - > \frac{\pi}{6}) \frac{2sinx + 1}{4cos2x - 1} = \\ = \frac{2sin( \frac{\pi}{6} )+ 1}{4cos( \frac{\pi}{3} ) - 1} = \frac{ 2\times 0.5 + 1}{4 \times 0.5 - 1} = \\ = \frac{2}{1} = 2" align="absmiddle" class="latex-formula">