Question

Zadanie w załączniku​

Answer 1

Odpowiedź:

Szczegółowe wyjaśnienie:

a.

[tex]\lim_{x \to 2} \frac{x^3-4}{2x-7}=[\frac{4}{-3}]=-1\frac{1}{3}[/tex]

b.

[tex]\lim_{x \to 3}\frac{9-x^2}{x^2-3x}=[\frac{0}{0} ]= \lim_{x \to 3} \frac{(3-x)(3+x)}{x(x-3)}= \lim_{x \to 3} \frac{3+x}{x} =[\frac{6}{3} ]=2[/tex]

c.

[tex]\lim_{x \to 1} \frac{x^3-1}{4x^2-7x+3}=[\frac{0}{0} ] = \lim_{x \to 1} \frac{(x-1)(x^2+x+1)}{4(x-1)(x-\frac{3}{4}) } = \lim_{x \to 1}\frac{x^2+x+1}{4x-3} =[\frac{3}{1} ]=3[/tex]

d.

[tex]\lim_{x \to \infty} \frac{3x^2+x-6}{x^2-x+5} = \lim_{x \to \infty}\frac{x^2(3+\frac{1}{x}-\frac{6}{x^2})} {x^2(1-\frac{1}{x} +\frac{5}{x^2} )} =[\frac{3}{1} ]=3[/tex]

e.

[tex]\lim_{x \to \infty} (2x^5-x^3+x-1)=[\infty-\infty+\infty-1]= \lim_{x \to \infty}x^5(2-\frac{1}{x^2} +\frac{1}{x^4} -\frac{1}{x^5}) =\\=[\infty*2]=\infty[/tex]