Question

limit(x->2) (x^3+3x^2-9x-2)/(x^3-x-6)

Answer 1

   
[tex]\displaystyle\\ \bold{\lim_{x \to 2} \frac{x^3+3x^2 -9x-2}{x^3 -x-6} } \\ \\ \texttt{Metoda 1 de rezolvare}\\\\ \texttt{Folosim L'hospital} \\ \\ \bold{\lim_{x \to 2} \frac{x^3+3x^2 -9x-2}{x^3 -x-6} =\lim_{x \to 2} \frac{(x^3+3x^2 -9x-2)'}{(x^3 -x-6)'} =} \\ \\ \bold{\lim_{x \to 2} \frac{3x^2+6x -9}{3x^2 -1} = \frac{3\cdot 4+6\cdot 2 -9}{3 \cdot4 -1} =\boxed{\frac{15}{11}} } \\ \\ \texttt{Metoda 2 de rezolvare}\\\\ \texttt{descompunem numaratorul si numitorul si simplificam.}\\\\ [/tex]


[tex]\displaystyle\\ \bold{\lim_{x \to 2} \frac{x^3+3x^2 -9x-2}{x^3 -x-6} = \lim_{x \to 2} \frac{\underbrace{x^3-2x^2}+\underbrace{5x^2 -10x}+\underbrace{x-2}}{\underbrace{x^3-2x^2} + \underbrace{2x^2 -4x}+\underbrace{3x-6}} =}\\\\ \bold{= \lim_{x \to 2} \frac{x^2(x-2)+5x(x-2) +(x-2)}{x^2(x-2)+2x(x-2)+3(x-2)}} = \\ \\ \bold{= \lim_{x \to 2} \frac{(x-2)(x^2+5x+1)}{(x-2)(x^2+2x+3) } =\lim_{x \to 2} \frac{x^2+5x+1}{x^2+2x+3 } } = \\ \\ \bold{= \frac{4+10+1}{4+4+3} = \boxed{\frac{15}{11} } }[/tex]