Question

O resultado do lim x→∞ 5x²-4x+3/3x-2 é:

a. +∞
b. Não existe
c. -∞
d. 5/3
e. -5/3

Answer 1

$L=\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{5x^{2}-4x+3}{3x-2}\\ \\ \\ =\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{5x^{2}\cdot\left(1-\frac{4x}{5x^{2}}+\frac{\infty}{5x^{2}} \right )}{3x\cdot (1-\frac{2}{3x})}\\ \\ \\ =\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{5x^{2}}{3x}\cdot \dfrac{1-\frac{4x}{5x^{2}}+\frac{3}{5x^{2}}}{1-\frac{2}{3x}}\\ \\ \\ =\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{5x}{3}\cdot \dfrac{1-\frac{4}{5x}+\frac{3}{5x^{2}}}{1-\frac{2}{3x}}~~~~~~\mathbf{(i)}$


Sabemos que

$\bullet~~\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{5x}{3}=+\infty~~~~\mathbf{(ii)}\\ \\ \\ \bullet~~\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{1-\frac{4}{5x}+\frac{3}{5x^{2}}}{1-\frac{2}{3x}}=1>0~~~~\mathbf{(iii)}$


Sendo assim, por $\mathbf{(ii)}$ e $\mathbf{(iii)},$ temos

$L=\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{5x}{3}\cdot \dfrac{1-\frac{4}{5x}+\frac{3}{5x^{2}}}{1-\frac{2}{3x}}=+\infty\\ \\ \\ \therefore~\boxed{\begin{array}{c} \underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{5x^{2}-4x+3}{3x-2}=+\infty \end{array}}$


Resposta: alternativa $\text{a. }+\infty.$