Question

Решите пж пределы!!!

Answer 1

$1)\ \ \lim\limits_{n \to \infty}\dfrac{n\, \sqrt[3]{5n^2}+\sqrt[4]{9n^8+1}}{(n+\sqrt{n})\sqrt{7-n+n^2} }=\Big[\ \dfrac{:n^2}{:n^2}\ \Big]=\lim\limits _{n \to \infty}\dfrac{\sqrt[3]{\frac{5}{n}}+\sqrt[4]{9+\frac{1}{n^8}}}{(\frac{1}{n}+\frac{1}{\sqrt{n^3}})\sqrt{\frac{7}{n^4}-\frac{1}{n^3}+\frac{1}{n^2}}}=\\\\\\=\dfrac{0+\sqrt[4]9}{0\cdot \sqrt0}=\infty$

$2)\ \ \lim\limits_{n \to \infty}\Big(\dfrac{n+1}{n-1}\Big)^{n}=\lim\limits_{n \to \infty}\Big(\Big(1+\dfrac{2}{n-1}\Big)^{\frac{n-1}{2}}\Big)^{\frac{2n}{n-1}}=e^{ \lim\limits _{n \to \infty}\frac{2n}{n-1}}=e^2$

$3)\ \ \lim\limits_{x \to 1}\dfrac{x^2-1}{lnx}=\lim\limits _{x \to 1}\dfrac{(x-1)(x+1)}{ln(1+(\underbrace {x-1}_{\alpha (x)}))}=\Big[\ ln(1+\alpha (x))\sim \alpha (x)\ ,\ \alpha (x)\to 0\ \Big]=\\\\\\=\lim\limits_{x \to 1}\dfrac{(x-1)(x+1)}{x-1}=\lim\limits_{x \to 1}(x+1)=1+1=2\\\\\\4)\ \ \lim\limits _{x \to -1}\dfrac{(x^3-2x-1)(x+1)}{x^4+4x^2-5}=\lim\limits _{x \to -1}\dfrac{(x+1)^2(x^2-x-1)}{(x-1)(x+1)(x^2+5)}=\\\\\\=\lim\limits_{x \to -1}\dfrac{(x+1)(x^2-x-1)}{(x-1)(x^2+5)}=\dfrac{0\cdot 1}{-2\cdot 6}=0$

$5)\ \ \lim\limits_{x \to 4}\dfrac{\sqrt{1+2x}-3}{\sqrt{x}-2}=\lim\limits_{x \to 4}\dfrac{(\sqrt{1+2x}-3)(\sqrt{1+2x}+3)(\sqrt{x}+2)}{(\sqrt{1+2x}+3)(\sqrt{x}-2)(\sqrt{x}+2)}=\\\\\\=\lim\limits_{x \to 4}\dfrac{(1+2x-9)(\sqrt{x}+2)}{(\sqrt{1+2}+3)(x-4)}=\lim\limits _{x \to 4}\dfrac{2(x-4)(\sqrt{x}+2)}{(\sqrt{1+2x}+3)(x-4)}=\\\\\\=\lim\limits _{x \to 4}\dfrac{2(\sqrt{x}+2)}{\sqrt{1+2x}+3}=\dfrac{2\cdot 4}{3+3}=\dfrac{8}{6}=\dfrac{4}{3}$

$6)\ \ \lim\limits_{x \to 0}\dfrac{ln(1+sinx)}{sin4x}=\Big[\ sin\alpha (x)\sim \alpha (x)\ ,\ ln(1+\alpha (x))\sim \alpha (x)\ ,\ \alpha (x)\to 0\ \Big]=\\\\\\=\lim\limits_{x \to 0}\dfrac{sinx}{4x}=\lim\limits_{x \to 0}\dfrac{x}{4x}=\dfrac{1}{4}$