• Matéria: Matemática
  • Autor: PPederiva
  • Perguntado 9 anos atrás

Considere a função:

f(x)= \frac{ x^{2} -5x+6}{x-2}

Qual é o valor de  \lim_{x \to \infty} f(x)?

Respostas

respondido por: Lukyo
1
\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{x^2-5x+6}{x-2}\\\\\\ =\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{x^2\cdot \left(1-\frac{5}{x}+\frac{6}{x^2}\right)}{x\cdot \left(1-\frac{2}{x} \right )}\\\\\\ =\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{x\cdot \diagup\!\!\!\! x\cdot \left(1-\frac{5}{x}+\frac{6}{x^2}\right)}{\diagup\!\!\!\! x\cdot \left(1-\frac{2}{x} \right )}\\\\\\ =\underset{x\to \infty}{\mathrm{\ell im}}~x\cdot \dfrac{1-\frac{5}{x}+\frac{6}{x^2}}{1-\frac{2}{x}}\\\\\\ =\underset{x\to \infty}{\mathrm{\ell im}}~g(x)\cdot h(x)


onde

g(x)=x~~\text{ e }~~h(x)=\dfrac{1-\frac{5}{x}+\frac{6}{x^2}}{1-\frac{2}{x}}

_________

Sabemos que

\underset{x\to \infty}{\mathrm{\ell im}}~g(x)\\\\ =\underset{x\to \infty}{\mathrm{\ell im}}~x=\infty


e que

\underset{x\to \infty}{\mathrm{\ell im}}~h(x)\\\\ =\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{1-\frac{5}{x}+\frac{6}{x^2}}{1-\frac{2}{x}}\\\\\\ =\dfrac{1-0+0}{1-0}=1~~~~~(1>0)


Logo, por propriedades operatórias de limites infinitos, temos que

\underset{x\to \infty}{\mathrm{\ell im}}~g(x)\cdot h(x)=\infty\\\\\\ \therefore~~\boxed{\begin{array}{c} \underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{x^2-5x+6}{x-2}=\infty \end{array}}


Bons estudos! :-)


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