Question

Please ,help me
I dont know how to solve it

Answer 1

1)
$\lim_{x \to 2} \frac{2- \sqrt{x+2} }{2-x} = \lim_{x \to 2} \frac{(2- \sqrt{x+2}) (2+ \sqrt{x+2})}{(2-x)(2+ \sqrt{x+2})} =\lim_{x \to 2} \frac{4-(x+2)}{(2-x)(2+ \sqrt{x+2})} =$
$=\lim_{x \to 2} \frac{2-x}{(2-x)(2+ \sqrt{x+2})} =\lim_{x \to 2} \frac{1}{2+ \sqrt{x+2}} = \frac{1}{2+ \sqrt{4} } = \frac{1}{4} =0.25$
2)
$\lim_{x \to9} \frac{3- \sqrt{x} }{x-9} = \lim_{x \to9} \frac{(3- \sqrt{x})(3+ \sqrt{x} )}{(x-9)(3+ \sqrt{x} )} = \lim_{x \to9} \frac{9-x}{(x-9)(3+ \sqrt{x} )} =$
$= \lim_{x \to9} -\frac{1}{3+ \sqrt{x}} =- \frac{1}{3+ \sqrt{9} } =- \frac{1}{6}$