Question

вычислите пределы, письменно от руки пожалуйста

Answer 1

$\lim_{x \to \ 3} \frac{x^2-9}{x-3} = \lim_{x \to \ 3} \frac{(x-3)(x+3)}{x-3} =\\= \lim_{x \to \ 3} (x+3)=3+3=6$

$\lim_{x \to \ 2} \frac{4b^2-x^2}{x-2b} = \lim_{x \to \ 2} \frac{(2b-x)(2b+x)}{x-2b} =\\= \lim_{x \to \ 2} -\frac{(2b-x)(2b+x)}{2b-x} = \lim_{x \to \ 2} -(2b+x)=-2b-2$

$\lim_{x \to \ -4} \frac{x^2+x-12}{x+4} = \lim_{x \to \ -4} \frac{(x+4)(x-3)}{x+4} =\\= \lim_{x \to \ -4} (x-3)=-4-3=-7$

$\lim_{x \to \ 5} \frac{x^2-4x-5}{x-5} = \lim_{x \to \ 5} \frac{(x+1)(x-5)}{x-5} =\\= \lim_{x \to \ 5} (x+1)=5+1=6$

$\lim_{x \to \ 3} \frac{x^3-27}{x-3} = \lim_{x \to \ 3} \frac{(x-3)(x^2+3x+9)}{x-3} =\\= \lim_{x \to \ 3} (x^2+3x+9)=3^2+3*3+9=27$

$\lim_{x \to \ 2} \frac{x-2}{x^2-4x+4} = \lim_{x \to \ 2} \frac{x-2}{(x-2)(x-2)} =\\= \lim_{x \to \ 2} \frac{1}{x-2} =\frac{1}{0}=+beskonechnost$