Question

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Lim
Все 6

Answer 1

$\displaystyle\\1)~\lim_{x\to 3}\dfrac{x^3-27}{x-3}=\lim_{x\to 3}\dfrac{\left(x-3\right)\left(x^2+3x+9\right)}{\left(x-3\right)}=\lim_{x\to 3}x^2+3x+9=9+9+9=\medskip\\=27\medskip\\2)~\lim_{x\to 3}\dfrac{4x^2-11x-3}{3x^2-8x-3}=\lim_{x\to 3}\dfrac{4\left(x+0.25\right)\left(x-3\right)}{3\left(x+\frac{1}{3}\right)\left(x-3\right)}=\lim_{x\to 3}\dfrac{4x+1}{3x+1}=\dfrac{12+1}{9+1}=1.3$

$\displaystyle\\3)~\lim_{x\to\infty}\dfrac{x^3+2x+1}{10x^3+x^2-80}=\lim_{x\to\infty}\dfrac{x^3\left(1+\frac{2}{x^2}+\frac{1}{x^3}\right)}{x^3\left(10+\frac{1}{x}-\frac{80}{x^3}\right)}=\lim_{x\to\infty}\dfrac{1+\frac{2}{x^2}+\frac{1}{x^3}}{10+\frac{1}{x}-\frac{80}{x^3}}=0.1$

$\displaystyle\\4)~\lim_{x\to 6}\dfrac{x-6}{3-\sqrt{x+3}}=\lim_{x\to 6}\dfrac{\left(x-6\right)\left(3+\sqrt{x+3}\right)}{\left(3-\sqrt{x+3}\right)\left(3+\sqrt{x+3}\right)}=\medskip\\=\lim_{x\to 6}\dfrac{\left(x-6\right)\left(3+\sqrt{x+3}\right)}{9-x-3}=\lim_{x\to 6}\dfrac{\left(x-6\right)\left(3+\sqrt{x+3}\right)}{-\left(x-6\right)}=\medskip\\=-\lim_{x\to 6}3+\sqrt{x+3}=-\left(3+\sqrt{6+3}\right)=-\left(3+3\right)=-6$

$\displaystyle\\5)~\lim_{x\to\infty}{\left(1+\dfrac{2}{3x}\right)}^x$

Пусть $t=\dfrac{2}{3x}$, тогда $x=\dfrac{2}{3t}$.

$\displaystyle\\t=\dfrac{2}{3x}\medskip\\x\to\infty\Rightarrow t\to 0\medskip\\\lim_{t\to 0}{\left(1+t\right)}^{\frac{2}{3t}}=\lim_{t\to 0}\left[\left(1+t\right)^{\frac{1}{t}}\right]^{\frac{2}{3}}=\left[\lim_{t\to 0}\left(1+t\right)^{\frac{1}{t}}\right]^{\frac{2}{3}}=e^\frac{2}{3}={\left(\sqrt[3]{e}\right)}^2$

$\displaystyle\\6)~\lim_{x\to 0}\dfrac{3x}{\sin 7x}$

Пусть $u=7x$, тогда $x=\dfrac{u}{7}$.

$\displaystyle\\u=7x\medskip\\x\to 0\Rightarrow u\to 0\medskip\\\lim_{u\to 0}\dfrac{\frac{3u}{7}}{\sin u}=\dfrac{3}{7}\lim_{u\to 0}\dfrac{u}{\sin u}=\dfrac{3}{7}\left[\lim_{u\to 0}\dfrac{\sin u}{u}\right]^{-1}=\dfrac{3}{7}\cdot 1^{-1}=\dfrac{3}{7}$