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$\displaystyle \lim_{x \to \frac{\pi}{6}} \frac{sin(x-\frac{\pi}{6})}{\frac{\sqrt3}{2}-cosx}=\frac{0}{0}=\lim_{x \to \frac{\pi}{6}}\frac{(sin(x-\frac{\pi}{6}))'}{(\frac{\sqrt3}{2}-cosx)'}=\\=\lim_{x \to \frac{\pi}{6}}\frac{cos(x-\frac{\pi}{6})}{sinx}=\frac{1}{\frac{1}{2}}=2$
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$\displaystyle \lim_{x \to \infty} (\frac{x+1}{x-2})^{2x-1}=(\frac{\infty}{\infty})^\infty=\lim_{x \to \infty} (\frac{x-2+3}{x-2})^{2x-1}=\\=[\lim_{x \to \infty} (1+\frac{1}{\frac{x-2}{3}})^\frac{x-2}{3}]^{\frac{3}{x-2}*(2x-1)}=e^{\displaystyle \lim_{x \to \infty}\frac{6x-3}{x-2}}=e^{\displaystyle \lim_{x \to \infty}\frac{x(6-\frac{3}{x})}{x(1-\frac{2}{x})}}=\\=e^6$
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$\displaystyle \lim_{x \to 0} \frac{\sqrt{1+x^2}-1}{x}=\frac{0}{0}= \lim_{x \to 0} \frac{\sqrt{1+x^2}-1}{x}*\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}+1}=\\=\lim_{x \to 0}\frac{x^2}{x(\sqrt{1+x^2}+1)}=\lim_{x \to 0}\frac{x}{(\sqrt{1+x^2}+1)}=0$