Question

( begin{aligned} lim _{n rightarrow-2} & frac{3-sqrt{1-4 x}}{cos sqrt{2-3 x}-sqrt{6-x}} &=frac{3-sqrt{1-4(-2)}}{sqrt{2-3(-2)-sqrt{6+2}}} &=frac{3-sqrt{9}}{sqrt{8}-sqrt{8}}>frac{3-3}{0}=frac{0}{0}=operatorname{lom} end{aligned} )
By D- L hesbital rule ( frac{lim _{x rightarrow-2}-frac{(-4)}{x sqrt{1-4 x}}}{frac{(-3)}{y sqrt{2-3 x}}+frac{1}{8 sqrt{6-x}}} frac{y / 1,}{sqrt{x-44}} )
( =frac{frac{4}{sqrt{1-4(-2)}}}{frac{-3}{sqrt{2-3(-2)}+frac{1}{sqrt{6}-(-2)}}}{frac{frac{4}{sqrt{9}}{frac{-3}{sqrt{8}}+frac{1}{sqrt{8}}}}-frac{frac{4}{sqrt{9}}}{frac{-2}{sqrt{8}}}-frac{4 sqrt{8}}{-2 sqrt{3}}}{frac{-2 sqrt{3}}{sqrt{9}}} )

# lim 3 - V1 – 4x a) Evaluate x-→-2 12 – 3x - 16 - X

Solution