Question

Помогите с буквой В, пожалуйста. Заранее спасибо

Answer 1

$1); ; limlimits _{x to 4}dfrac{sqrt{x+5}-3}{x-4}=limlimits _{x to 4}dfrac{(sqrt{x+5}-3)(sqrt{x+5}+3)}{(x-4)(sqrt{x+5}+3)}=\\\=limlimits _{x to 4}dfrac{(x+5)-9}{(x-4)(sqrt{x+5}+3)}= limlimits _{x to 4}dfrac{x-4}{(x-4)(sqrt{x+5}+3)}= limlimits _{x to 4}dfrac{1}{sqrt{x+5}+3}=\\\=dfrac{1}{3+3}=dfrac{1}{6}$

$2); ; limlimits _{x to 2}dfrac{x-2}{sqrt{x+2}-2}=limlimits _{x to 2}dfrac{(x-2)(sqrt{x+2}+2)}{(x+2)-4}=limlimits _{x to 2}(sqrt{x+2}+2)=4\\\3); ; limlimits _{x to 1}dfrac{2sqrt{x+3}-4}{x-1}=limlimits _{x to 1}dfrac{4(x+3)-16}{(x-1)(2sqrt{x+3}+4)}=\\\=limlimits _{x to 1}dfrac{4, (x-1)}{(x-1)(2sqrt{x+3}+4)}=limlimits _{x to 1}dfrac{4}{2sqrt{x+3}+4}=dfrac{4}{2cdot 2+4}=dfrac{1}{2}$

$4); ; limlimits _{x to 3}dfrac{x-3}{sqrt{3x+7}-4}=limlimits _{x to 3}dfrac{(x-3)(sqrt{3x+7}+4)}{(3x+7)-16}=limlimits _{x to 3}dfrac{(x-3)(sqrt{3x+7}+4)}{3x-9}=\\\=limlimits _{x to 3}dfrac{(x-3)(sqrt{3x+7}+4)}{3(x-3)}=limlimits_{x to 3}dfrac{sqrt{3x+7}-4}{3}=dfrac{4-4}{3}=0$