Question

Помогите пожалуйста

$\lim_{x \to \00} \frac{x+ x^{2} }{ \sqrt{1+3x}- \sqrt{1-2x} } \lim_{x \to \00} \frac{1-cos3x}{7x^2} \lim_{x \to \infty} ( \frac{2x-1}{2x+1} )^{x-1} \lim_{x \to \ \frac{ \pi }{2} } \frac{2-2cos( \frac{ \pi }{2} -x)}{ ( \frac{ \pi }{2}) ^{2} } \lim_{x \to \infty} ( \frac{2 x^{2} +3x+1}{2 x^{2} -1}) ^{ x^{2} }$

Answer 1

$lim_{\to 0}\frac{x+x^2}{\sqrt{1+3x}-\sqrt{1-2x}}=lim\frac{x(x+1)(\sqrt{1+3x}+\sqrt{1-2x})}{(1+3x)-(1-2x)}=\\\\=lim\frac{x(x+1)(\sqrt{1+3x}+\sqrt{1-2x})}{5x}=\frac{1}{5}lim_{x\to 0}(x+1)(\sqrt{1+3x}+\sqrt{1-2x})=\\\\=\frac{1}{5}\cdot (1+1)=\frac{2}{5}$

$2)\; lim_{x\to 0}\frac{1-cos3x}{7x^2}=lim\frac{2sin^2\frac{3x}{2}}{7x^2}=lim_{x\to 0}\frac{2\cdot (\frac{3x}{2})^2}{7x^2}=\frac{9}{14}\\\\3)\; lim_{x\to \infty}(\frac{2x-1}{2x+1})^{x-1}=lim((1+\frac{-2}{2x+1})^{\frac{2x+1}{-2}})^{\frac{-2(x-1)}{2x+1}}=\\\\=lim_{x\to \infty}e^{\frac{-2x+1}{2x+1}}=e^{-1}$

$4)\; lim_{x\to \frac{\pi}{2}}\frac{2-2cos(\frac{\pi}{2}-x)}{(\frac{\pi}{2})^2}=[\frac{2-2\cdot cos0}{(\frac{\pi}{2})^2}]=[\frac{0}{(\frac{\pi}{2})^2}]=0$

$5)\; lim_{x\to \infty}(\frac{2x^2+3x+1}{2x^2-1})^{x^2}=lim((1+\frac{3x+2}{2x^2-1})^{\frac{2x^2-1}{3x+2}})^\, {\frac{(3x+2)x^2}{2x^2-1}}}=\\\\=lim_{x\to \infty}\, e^{\frac{3x^3+2x^2}{2x^2-1}}=e^{\infty}= \left \{ {{0,\; esli\; x\to -\infty} \atop {+\infty,\; esli\; x\to +\infty}} \right.$