Question

Помогите с буквой В, пожалуйста. Заранее спасибо

Answer 1

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

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

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