Question

Помогите пожалуйста Пределы

Answer 1

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