Question

Como resolvo esse limite

a) -∞
b) +∞
c) Nao existe
d) 5/3
e)-5/2

Answer 1

$\lim_{x \to \infty} \frac{5x^2-4x+3}{3x-2}= \lim_{x \to \infty} \frac{x^2}{x} \left[\begin{array}{ccc} \frac{5- \frac{4}{x}+ \frac{3}{x^2} }{3- \frac{2}{x} } \end{array}\right]$
Constante dividido por ${ \infty}$ é igual e $0$
Logo, sobra: $\lim_{x \to \infty} \frac{x^2}{x} \left[\begin{array}{ccc} \frac{5}{3} \end{array}\right]= \lim_{x \to \infty} x \left[\begin{array}{ccc} \frac{5}{3} \end{array}\right]= \frac{5{\infty}}{3} = \frac{{\infty}}{3}={\infty}$

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