a=1;
b=0;
[tex]\displaystyle&\lim_{n\to\infty}\left(an-\sqrt{n^2+bn-2}\right)\cdot n=1\\[/tex]
"Rationalizam" numaratorul
(inmultim si impartim paranteza la [tex]an+\sqrt{n^2+bn-2}[/tex] )
[tex]\displaystyle&\lim_{n\to\infty}\dfrac{\left(an-\sqrt{n^2+bn-2}\right)\left(an+\sqrt{n^2+bn-2}\right)}{an+\sqrt{n^2+bn-2}}\cdot n=1\\[/tex]
[tex]\displaystyle&\lim_{n\to\infty}\dfrac{(an)^2-(n^2+bn-2)}{an+\sqrt{n^2+bn-2}}\cdot n=1\\[/tex]
[tex]\displaystyle&\lim_{n\to\infty}\dfrac{n^2\left(a^2-1\right)-bn+2}{an+\sqrt{n^2+bn-2}}\cdot n=1\\[/tex]
[tex]\displaystyle&\lim_{n\to\infty}\cfrac{n^2\left(a^2-1\right)-bn+2}{n\left(a+\sqrt{1+b\cfrac1n-2\cfrac{1}{n^2}}\right)}\cdot n=1\\[/tex]
[tex]\displaystyle&\lim_{n\to\infty}\cfrac{n^2\left(a^2-1\right)-bn+2}{a+\sqrt{1+b\cfrac1n-2\cfrac{1}{n^2}}}=1\\[/tex]
Observam ca numitorul are limita, deci si numaratorul trebuie sa aiba limita. Atunci termenii cu [tex]n[/tex] si [tex]n^2[/tex] trebuie sa se reduca. Mai observam si ca limita numitorului trebuie sa fie diferita de 0. Astfel:
[tex]\begin{equation*}\left\{ \begin{aligned} &a^2-1=0\\&-b=0\\&\lim_{n\to\infty}\left(a+\sqrt{1+b\underset{\to0}{\cfrac1n}-2\underset{\to0}{\cfrac{1}{n^2}}}\right)\not=0\end{aligned}\right.\iff \left\{ \begin{aligned} &a^2=1\\&b=0\\&a+\sqrt{1}\not=0\end{aligned}\right.\iff\end{equation*}[/tex]
[tex]\left\{ \begin{aligned} &a=\pm\sqrt1\\&b=0\\&a+1\not=0\end{aligned}\right. \iff \left\{ \begin{aligned} &a=\pm1\\&b=0\\&a\not=-1\end{aligned}\right.\implies \left\{\begin{aligned}&a=1\\&b=0\end{aligned}\right.[/tex]
atunci:
[tex]\displaystyle&\lim_{n\to\infty}\cfrac{n^2\left(a^2-1\right)-bn+2}{a+\sqrt{1+b\cfrac1n-2\cfrac{1}{n^2}}}=\cfrac{\displaystyle\lim_{n\to\infty}\left(n^2\left(a^2-1\right)-bn+2\right)}{\displaystyle\lim_{n\to\infty}\left(a+\sqrt{1+b\cfrac1n-2\cfrac{1}{n^2}}\right)}=1\\\cfrac{\displaystyle\lim_{n\to\infty}\left(n^2\left(1^2-1\right)-0\cdot n+2\right)}{\displaystyle\lim_{n\to\infty}\left(1+\sqrt{1+0\cdot\cfrac1n-2\cfrac{1}{n^2}}\right)}=1\\[/tex]
[tex]\cfrac{\displaystyle\lim_{n\to\infty}2}{\displaystyle\lim_{n\to\infty}\left(1+\sqrt{1-2\underset{\to0}{\cfrac{1}{n^2}}}\right)}=1 \implies\\\dfrac{2}{1+\sqrt1}=1\\\dfrac{2}{2}=1 \text{(adevarat)}[/tex]
Deci, solutiile:
a=1;
b=0;
Sunt bune
