Question

1)$lim_{x to x0} frac{2- sqrt{2x+4} }{x}$
2)$lim_{x to - -2} frac{2x+ x^{2} }{ x^{2} +5x+6}$

Answer 1

$displaystyle lim_{x to 0} frac{2-sqrt{2x+4}}{x}= lim_{x to 0} frac{(2-sqrt{2x+4})(2+sqrt{2x+4})}{x(2+sqrt{2x+4})}=\\=lim_{x to 0} frac{4-(2x+4)}{x(2+sqrt{2x+4})}=lim_{x to 0} frac{-2x}{x(2+sqrt{2x+4})}=\\=lim_{x to 0} frac{-2}{(2+sqrt{2x+4})}=frac{-2}{2+sqrt4}=frac{-2}{2+2}=-frac{1}2$

$displaystyle lim_{n to -2} frac{2x+x^2}{x^2+5x+6}= left[begin{array}{ccc}x^2+5x+6=0\D=25-4*6=25-24=1\x_1=frac{-5+1}{2}=-2\x_2=frac{-5-1}2=-3end{array}right] =\\\ lim_{n to -2} frac{x(x+2)}{(x+2)(x+3)}=lim_{n to -2}frac{x}{x+3}=frac{-2}{-2+3}=-2$