Question

o valor do lim (x tende - infinito) 2x²-5x+1/4x²+3x-7?

Answer 1

$\underset{x\to-\infty}{lim}\frac{2x^2-5x+1}{4x^2+3x-7}=\underset{x\to-\infty}{lim}\frac{2x^2-5x+1}{4x^2+3x-7}\div\frac{x^2}{x^2}=\underset{x\to-\infty}{lim}\frac{2-\frac5x+\frac{1}{x^2}}{4+\frac3x-\frac{7}{x^2}}=$ $\frac{2-0+0}{4+0-0}=\frac24=\boxed{\frac12}$