Question

помогите пожалуйста с матешой

Answer 1

$\lim_{x \to \infty}\frac{100x+3x^2}{x^2-11}= \lim_{x \to \infty}\frac{\frac{100x}{x^2} +\frac{3x^2}{x^2}}{\frac{x^2}{x^2}-\frac{11}{x^2}}= \lim_{x \to \infty}\frac{\frac{100}{x}+3}{1-\frac{11}{x^2}}=\frac{0+3}{1-0}=\frac{3}{1}=3$

$\lim_{x \to 4} \frac{x^2-11x+28}{x^2-5x+4}=\lim_{x \to 4} {\frac{(x-4)(x-7)}{(x-4)(x-1)}=\lim_{x \to 4} {\frac{x-7}{x-1}=\frac{4-7}{4-1}=\frac{-3}{3}=-1$

$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{4x}=\lim_{x \to 0}\frac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{4x(\sqrt{1+x}+\sqrt{1-x})}}=\\\\=\lim_{x \to 0}\frac{(\sqrt{1+x})^2-(\sqrt{1-x})^2}{4x(\sqrt{1+x}+\sqrt{1-x})}=\lim_{x \to 0}\frac{(1+x)-(1-x)}{4x(\sqrt{1+x}+\sqrt{1-x})}=\\\\=\lim_{x \to 0}\frac{1+x-1+x}{4x(\sqrt{1+x}+\sqrt{1-x})}=\lim_{x \to 0}\frac{2x}{4x(\sqrt{1+x}+\sqrt{1-x})}=\\\\=\lim_{x \to 0}\frac{1}{2(\sqrt{1+x}+\sqrt{1-x})}=\frac{1}{2(\sqrt{1+0}+\sqrt{1-0)}}}=\frac{1}{2(1+1)}=$

$=\frac{1}{2*2}=\frac{1}{4}=0,25$

$\lim_{x \to 0}\frac{cos6x-cos4x}{x^2}= \lim_{x \to 0}\frac{-2sin5x*sinx}{x^2}=\lim_{x \to 0}(-2*5*\frac{sin5x}{5x}*\frac{sinx}{x}) =\\\\=-10\lim_{x \to 0}\frac{sin5x}{5x}*\lim_{x \to 0}\frac{sinx}{x}=-10*1*1=-10$

$\lim_{x \to \infty}(\frac{x-3}{x})^{\frac{x}{3}}= \lim_{x \to \infty}(1+\frac{-3}{x})^{\frac{x}{-3}*(-1)}=\lim_{x \to \infty}((1+\frac{-3}{x})^{\frac{x}{-3}})^{-1}=\\\\= \lim_{x \to \infty}e^{-1}=e^{-1}=\frac{1}{e}$