Доказателство

[Гост]

IMG_20220204_205548_edit_160391103773962.jpg (862.13 KiB) Прегледано 474 пъти

[KOPMOPAH]

$$\lim_{x \to +\infty}\left(\frac{x^3}{2x^2-1}-\frac{x^2}{2x+1}\right)=\lim_{x \to +\infty}\left(\frac{x^3(2x+1)-x^2(2x^2-1)}{(2x^2-1)(2x+1)}\right)=\lim_{x \to +\infty}\left(\frac{\cancel{2x^4}+x^3-\cancel{2x^4}+x^2}{4x^3+2x^2-2x-1}\right)=$$ $$=\lim_{x \to +\infty}\left(\frac{\cancel{x^3}(1+0)}{\cancel{x^3}(4+0-0-0)}\right)=\frac 14$$

[Гост]

Благодаря!!!

[KOPMOPAH]

В общия случай се умножава по сумата на радикалите$$\lim_{x \to \pm \infty}\left(\sqrt{x^2+ax+b}-\sqrt{x^2+cx+d}\right)=\lim_{x \to \pm \infty}\left(\frac{\left(\sqrt{x^2+ax+b}-\sqrt{x^2+cx+d}\right)\left(\sqrt{x^2+ax+b}+\sqrt{x^2+cx+d}\right)}{\sqrt{x^2+ax+b}+\sqrt{x^2+cx+d}}\right)=$$ $$=\lim_{x \to \pm \infty}\left(\frac{\cancel{x^2}+ax+b-\cancel{x^2}-cx-d}{\sqrt{x^2+ax+b}+\sqrt{x^2+cx+d}}\right)=\lim_{x \to \pm \infty}\left(\frac{\cancel{x}(a+\displaystyle\frac bx-c-\frac dx)}{\cancel x\left(\sqrt{1+\displaystyle{\frac ax}+ \frac b{x^2}}+\sqrt{1+\displaystyle\frac cx+\frac d{x^2}}\right)}\right)=\left(\frac{(a+0-c-0)}{\left(\sqrt{1+0+ 0}+\sqrt{1+0+0}\right)}\right)=\frac {a-c}2$$

За подточка в) замести с подходящите стойности.

[nikola.topalov]

...$$\lim_{x \to \pm \infty}\left(\frac{\cancel{x^2}+ax+b-\cancel{x^2}-cx-d}{\sqrt{x^2+ax+b}+\sqrt{x^2+cx+d}}\right)=\lim_{x \to \pm \infty}\left(\frac{\cancel{x}\left(a+\displaystyle\frac bx-c-\frac dx\right)}{\cancel x\left(\sqrt{1+\displaystyle{\frac ax}+ \frac b{x^2}}+\sqrt{1+\displaystyle\frac cx+\frac d{x^2}}\right)}\right)$$...

Всъщност $$\lim_{x \to \pm \infty}\frac{ax+b-cx-d}{\sqrt{x^2+ax+b}+\sqrt{x^2+cx+d}}=\lim_{x \to \pm \infty}\frac{x\left(a+\displaystyle\frac bx-c-\frac dx\right)}{|x|\left(\sqrt{1+\displaystyle{\frac ax}+ \frac b{x^2}}+\sqrt{1+\displaystyle\frac cx+\frac d{x^2}}\right)}$$ и при [tex]x\to-\infty[/tex] имаме [tex]|x|=-x[/tex]. Следователно $$\lim_{x \to \pm \infty}\left(\sqrt{x^2+ax+b}-\sqrt{x^2+cx+d}\right)=\pm \frac {a-c}2$$

[KOPMOPAH]

Така е!