7 przykładów, zadania w załączniku- analiza matematyczna.... Question from @derrapes

October 2018 1 101 Report

7 przykładów, zadania w załączniku- analiza matematyczna

$\\\lim_{n\to\infty}\frac{-3n^2-2}{6-4n^3}=\\ \lim_{n\to\infty}\frac{n^2(-3-\frac{2}{n^2})}{n^3(\frac{6}{n^3}-4)}=\\ \lim_{n\to\infty}\frac{1}{x}\cdot\frac{-3-\frac{2}{n^2}}{\frac{6}{n^3}-4}=\\ 0\cdot\frac{-3-0}{0-4}=0$

$\\\lim_{x\to0}\frac{\sin 3x}{10x}=\\ \lim_{x\to0}\frac{1}{10}\cdot\frac{\sin 3x}{x}=\\ \lim_{x\to0}\frac{1}{10}\cdot\frac{(\sin 3x)'}{x'}=\\ \lim_{x\to0}\frac{1}{10}\cdot3\cos3x=\\ \lim_{x\to0}\frac{3}{10}\cdot\cos3x=\\ \frac{3}{10}\cdot\cos0=\\ \frac{3}{10}\cdot1=\\ \frac{3}{10}$

2.
$\\f(x)=\sqrt{4x^3-2x^2}\\ f'(x)=\frac{1}{2\sqrt{4x^3-2x^2}}\cdot(12x^2-4x)\\ f'(x)=\frac{6x^2-2x}{\sqrt{4x^3-2x^2}}\\ f'(x)=\frac{2x(3x-1)}{\sqrt{x^2(4x-2)}}\\ f'(x)=\frac{2x(3x-1)}{|x|\sqrt{4x-2}}\\$

3.
4x²-1≠0
4x²≠1
x²≠1/4
x≠-1/2 ∨ x≠1/2

asymptota pionowa:
$> ⇒ asymptota pionowa x=1/2 <img src=$ ⇒ brak asymptot poziomych

asymptota ukośna:

$\\\lim_{x\to\pm\infty}\frac{\frac{1-x^3}{4x^2-1}}{x}=\\ \lim_{x\to\pm\infty}\frac{1-x^3}{4x^3-x}=\\ \lim_{x\to\pm\infty}\frac{x^3(\frac{1}{x^3}-1)}{x^3(4-\frac{1}{x^2})}=\\ \lim_{x\to\pm\infty}\frac{\frac{1}{x^3}-1}{4-\frac{1}{x^2}}=\\ \frac{0-1}{4-0}=\\ -\frac{1}{4}\\\\ \lim_{x\to\pm\infty}\frac{1-x^3}{4x^2-1}\ plus\ \frac{1}{4}x=\\ \lim_{x\to\pm\infty}\Big(\frac{1-x^3}{4x^2-1}\ plus\ \frac{1}{4}x\Big)=\\ \lim_{x\to\pm\infty}\frac{4(1-x^3)+x(4x^2-1)}{4(4x^2-1)}=\\ \lim_{x\to\pm\infty}\frac{4-4x^3+4x^3-x}{16x^2-4}=\\ \lim_{x\to\pm\infty}\frac{4-x}{16x^2-4}=\\ \lim_{x\to\pm\infty}\frac{x(\frac{4}{x}-1)}{x^2(16-\frac{4}{x^2})}=\\ \lim_{x\to\pm\infty}\frac{1}{x}\cdot\frac{\frac{4}{x}-1}{16-\frac{4}{x^2}}=\\ 0\cdot\frac{0-1}{16-0}= 0$