Question

lim x-->4 x²-6x+8 / x2-16

Answer 1

Объяснение:

1.

$\lim_{x \to 4} \frac{x^2-6x+8}{x^2-16} .\\ \frac{x^2-6x+8}{x^2-16} =\frac{x^2-4x-2x+8}{x^2-4^2}=\frac{x*(x-4)-2*(x-4)}{(x-4)*(x+4)}=\frac{(x-4)*(x-2)}{(x-4)*(x+4)}=\frac{x-2}{x+4} .\ \ \ \ \Rightarrow\\ \lim_{x \to 4} \frac{x-2}{x+4}=\frac{4-2}{4+4}=\frac{2}{8}=\frac{1}{4}.$

2.

$\lim_{x \to 4}\frac{x^2-6x+8}{x^2-16}= \lim_{x \to 4}\frac{(x^2-6x+8)'}{(x^2-16)'}= \lim_{x \to 4} \frac{2x-6}{2x}=\frac{2*4-6}{2*4} =\frac{8-6}{8}=\frac{2}{8}=\frac{1}{4} .$