Question

, решить выделенные пределы

Answer 1

$1)\ \ \lim\limits _{x \to 4}\dfrac{5-\sqrt{21+x}}{\sqrt{13-x}-3}=\lim\limits _{x \to 4}\dfrac{(5-\sqrt{21+x})(5+\sqrt{21+x})(\sqrt{13-x}+3)}{(5+\sqrt{21+x})(\sqrt{13-x}-3)(\sqrt{13-x}+3)}=\\\\\\=\lim\limits _{x \to 4}\dfrac{(25-(21+x))(\sqrt{13-x}+3)}{(5+\sqrt{21+x})(13-x-9)}=\lim\limits _{x \to 4}\dfrac{(4-x))(\sqrt{13-x}+3)}{(5+\sqrt{21+x})(4-x)}=\\\\\\=\lim\limits _{x \to 4}\dfrac{\sqrt{13-x}+3}{5+\sqrt{21+x}}=\dfrac{3+3}{5+5}=\dfrac{9}{10}=0,9$

$2)\ \ \lim\limits _{x \to 0}\dfrac{4\cdot tg8x\cdot cos3x}{3\cdot cos7x\cdot sin6x}=\lim\limits _{x \to 0}\dfrac{4\cdot 8x\cdot 1}{3\cdot 1\cdot 6x}=\dfrac{4\cdot 8}{3\cdot 6}=\dfrac{16}{9}\\\\\\\star \ \ tg8x\sim 8x\ \ ,\ \ \ sin6x\sim 6x\ \ ,\ \ esli\ \ x\to 0\ \ \star$