Oblicz granicę (chodzi o zadanie 3b; 3c i całe czwarte baaaardzo pilne !!!!

3. b) [latex]\ lim_{x o 5} dfrac{3x^2-14x-5}{2x^2-11x+5} =lim_{x o 5} dfrac{3x^2-15x+x-5}{2x^2-10x-x+5} = \ \lim_{x o 5} dfrac{3x(x-5)+(x-5)}{2x(x-5)-(x-5)} =lim_{x o 5} dfrac{(3x+1)(x-5)}{(2x-1)(x-5)} = \ \lim_{x o 5} dfrac{3x+1}{2x-1} = dfrac{3cdot5+1}{2cdot5-1} = dfrac{16}{9} \c) \ \lim_{x o -2} dfrac{-x^3-4x^2+4x+16}{x^3+4x^2+x-6} =lim_{x o -2} dfrac{-x^2(x+4)+4(x+4)}{x^3+4x^2+x-6} = \ \lim_{x o -2} dfrac{(x+4)(4-x^2)}{x^2(x+2)+2x(x+2)-3(x+2)} = [/latex] [latex]\lim_{x o -2} dfrac{(x+4)(2-x)(x+2)}{(x+2)(x^2+2x-3)} =lim_{x o -2} dfrac{(x+4)(2-x)}{x^2+2x-3} = \ \ dfrac{(-2+4)(2+2)}{4-4-3} = -dfrac{8}{3} [/latex] 4. a) [latex]\ lim_{x o0} dfrac{1- sqrt{x+1} }{x } dfrac{[H]}{=} lim_{x o0} dfrac{- frac{1}{2sqrt{x+1} }}{1 } =- dfrac{1}{2} \b) \ lim_{x o 1} dfrac{1-x}{2- sqrt{2x+2} } dfrac{[H]}{=} lim_{x o 1} dfrac{-1}{- frac{ 2}{2sqrt{2x+2}} } = dfrac{2cdot2}{2} =2 \c) \ lim_{x o 0} dfrac{x}{1- sqrt{1+x} } dfrac{[H]}{=} lim_{x o 0} dfrac{1}{- frac{1}{ 2sqrt{1+x}} }=- dfrac{2}{1} =-2 [/latex] d) [latex]\ lim_{x o 1} dfrac{1-x}{2(1- sqrt[3]{x} )} dfrac{[H]}{=} lim_{x o 1} dfrac{-1}{2(- frac{1}{3} x^{- frac{2}{3}})} = dfrac{1}{ frac{2}{3} } = dfrac{3}{2} [/latex]