Задачи с основна граница

[StarWars2015]

Здравейте! Ще бъда изключително благодарен, ако ми обясните как се решават въпросните задачи, използвайки основната граница. Благодаря предварително!

[Добромир Глухаров]

$\lim_{n\to\infty}\left(\frac{n^2-5n+6}{n^2+4n+5}\right)^{2n-1}=\lim_{n\to\infty}\left(\frac{n^2+4n+5-9n+1}{n^2+4n+5}\right)^{2n-1}=\lim_{n\to\infty}\left(1+\frac{1-9n}{n^2+4n+5}\right)^{2n-1}=\lim_{n\to\infty}\left[\left(1+\frac{1-9n}{n^2+4n+5}\right)^{\frac{n^2+4n+5}{1-9n}}\right]^{\frac{(1-9n)(2n-1)}{n^2+4n+5}}$

$\lim_{n\to\infty}\frac{1-9n}{n^2+4n+5}=\lim_{n\to\infty}\frac{1}{n}\cdot\frac{\frac{1}{n}-9}{1+\frac{4}{n}+\frac{5}{n^2}}=0.(-9)=0$

$\Rightarrow\lim_{n\to\infty}\left[\left(1+\frac{1-9n}{n^2+4n+5}\right)^{\frac{n^2+4n+5}{1-9n}}\right]^{\frac{(1-9n)(2n-1)}{n^2+4n+5}}=e^{\lim_{n\to\infty}\frac{(1-9n)(2n-1)}{n^2+4n+5}}=e^{\lim_{n\to\infty}\frac{2n-1-18n^2+9n}{n^2+4n+5}}=e^{\lim_{n\to\infty}\frac{-18n^2+11n+1}{n^2+4n+5}}=e^{\lim_{n\to\infty}\frac{-18+\frac{11}{n}+\frac{1}{n^2}}{1+\frac{4}{n}+\frac{5}{n^2}}}=e^{-18}$

[Добромир Глухаров]

$\lim_{n\to\infty}\frac{n^4+n.2^n+1}{n^3+3^n+1}=\lim_{n\to\infty}\frac{n+\frac{2^n}{n^2}+\frac{1}{n^3}}{1+\frac{3^n}{n^3}+\frac{1}{n^3}}=\lim_{n\to\infty}\frac{\frac{n}{2^n}+\frac{1}{n^2}+\frac{1}{n^3.2^n}}{\frac{1}{2^n}+\frac{3^n}{n^3.2^n}+\frac{1}{n^3.2^n}}=\lim_{n\to\infty}\frac{n\left(\frac{1}{2}\right)^n+0+0}{0+\frac{\left(\frac{3}{2}\right)^n}{n^3}+0}=\frac{0}{\infty}=0$

Използва се, че $\lim_{n\to\infty}\frac{a^n}{n^k}=\begin{cases}0;\ 0<a<1\\\infty;\ a>1\end{cases}$

[Добромир Глухаров]

$\lim_{x\to+\infty}\frac{e^{x^2}-cos^2x}{x^2}=\lim_{x\to+\infty}\frac{e^{x^2}}{x^2}-\lim_{x\to+\infty}\left(\frac{cos\ x}{x}\right)^2=+\infty-0^2=+\infty$

Използва се, че $\lim_{u\to+\infty}\frac{a^u}{u}=+\infty$ за $a>1$ и $\lim_{x\to+\infty}\frac{u(x)}{x}=0$ за $|u(x)|\le c<+\infty$

[Добромир Глухаров]

$x\to 0\Rightarrow e^{x^2}\sim 1+x^2;\ cos^2x=1-sin^2x\sim 1-x^2$

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

[StarWars2015]

Благодаря!