Qual o limite da função f(x) = (2x² + 6x + 4)/(2x+2) quando x se aproxima de -2?

\[2x^2+6x+4\,\,| \underline{\,\,2x+2}\]

1. Dividindo \(2x^2\)por \(2x\) tem-se o seguinte:

\[\eqalign{ 2x^2+6x+4\,\,| &\underline{\,\,2x+2} \\ &\,\,\,\, \color{Blue}{x} }\]

Multiplicando \(x\)por \(2x+2\)e realizando a subtração, tem-se o seguinte:

\[\eqalign{&\,\,\,\,\,\,\,2x^2+6x+4\,\,| \underline{\,\,2x+2} \\ &\underline{\color{Blue}{-(2x^2+2x)} }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\color{Black}x \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\color{Blue}{4x+4} }\]

1. Dividindo \(4x\)por \(2x\) tem-se o seguinte:

\[\eqalign{&\,\,\,\,\,\,\,2x^2+6x+4\,\,| \underline{\,\,2x+2} \\ &\underline{{-(2x^2+2x)} }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \color{Blue}{+2} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{4x+4} }\]

Multiplicando \(2\)por \(2x+2\)e realizando a subtração, tem-se o seguinte:

\[\eqalign{&\,\,\,\,\,\,\,2x^2+6x+4\,\,| \underline{\,\,2x+2} \\ &\underline{{-(2x^2+2x)} }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x+2 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{4x+4} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\color{Blue}{\underline{-(4x+4)} } \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\color{Blue}{0} }\]

Com quociente \(Q=x+2\)e resto \(R=0\) o resultado da divisão \((2x^2+6x+4)/(2x+2)\)é:

\[\eqalign{ 2x^2+6x+4 &= (2x+2)\cdot Q+R \\ &= (2x+2)\cdot(x+2) +0 \\ {2x^2+6x+4 \over 2x+2} &= x+2 \\ }\]

Portanto, a função \(f(x) = (2x^2 + 6x + 4)/(2x+2)\)é:

\[f(x)=x+2\]

Portanto, quando \(x\)se aproxima a \(-2\) o limite de \(f(x)\)é:

\[\eqalign{ \lim_{x \to -2} f(x) &= \lim_{x \to -2} (x+2) \\ &= -2+2 \\ &= 0 }\]

Concluindo, quando \(x\)se aproxima de \(-2\) o limite de \(f(x) = (2x^2 + 6x + 4)/(2x+2)\)é \(\boxed{\lim_{x \to -2} f(x)=0}\)