Само

за да освободим място в страниците на теми без отговор.
I група
1. а) [tex]\lim_{x \to 2} \frac{5\cancel{(x-2)}^1}{\cancel{(x-2)}(x+1) }=\frac{5}{2+1}=\frac{5.1}{3}[/tex]
1. б) [tex]\lim_{x \to \infty }\frac {\cancel{x^2}\left ( 2-\frac{10}{x } \right )}{\cancel{x^2}\left ( \frac{3}{x }-7 \right ) }=-\frac{2}{ 7}[/tex]
1. в) [tex]\lim_{x \to \infty }\frac {\cancel{x^2}\left ( \frac{1}{x }+ \frac{2}{x^2 } \right )}{{\cancel{x^2}\left (1+ \frac{4}{x }+\frac{5}{x^2 } \right )}}=\frac{0+0}{1+0+0}=0[/tex]
1. г) [tex]\lim_{x \to 0} \frac{3sin3x}{x}=9\lim_{3x \to 0} \frac{sin3x}{3x}=9.1=9[/tex]
2.а) [tex]y=\frac{x^2-3x+5}{x+1 }=\frac{(x-4)(x+1)+9}{x+1}[/tex]
[tex]y=x-4+\frac{9}{ x+1}[/tex]
[tex]y'=1-\frac{9}{(x+1)^2 }[/tex]
[tex]y''=\frac{18}{(x+1)^3 }[/tex]
2.б) [tex]y=sin^25x[/tex]
[tex]y'=2sin5x.cos5x.5=5sin10x[/tex]
[tex]y''=5cos10x.10=50cos10x[/tex]
Feci, quod potui, faciant meliora p0tentes.
Сторих каквото можах, по-добрите по-добро да направят.