Question

Calcule o limite no infinito

     $\lim\limits_{x\to \infty}\dfrac{5^x-3^x+1}{5^x+3^x+\frac{1}{x}}$

sem recorrer às regras de L'Hôpital.

Answer 1

É dado o limite:

$L=\lim_{x\to\infty}\dfrac{5^x-3^x+1}{5^x+3^x+\frac{1}{x}}$

Vamos dividir o numerador e o denominador por $5^x$:

$L=\lim_{x\to\infty}\dfrac{5^x-3^x+1}{5^x+3^x+\frac{1}{x}}\\\\ L=\lim_{x\to\infty}\dfrac{\dfrac{5^x-3^x+1}{5^x}}{~~\dfrac{5^x+3^x+\frac{1}{x}}{5^x}~~}\\\\ L=\lim_{x\to\infty}\dfrac{\dfrac{5^x}{5^x}-\dfrac{3^x}{5^x}+\dfrac{1}{5^x}}{~~\dfrac{5^x}{5^x}+\dfrac{3^x}{5^x}+\dfrac{1}{x\cdot {5^x}}~~}\\\\ L=\lim_{x\to\infty}\dfrac{1-\left(\dfrac{3}{5}\right)^x+\dfrac{1}{5^x}}{~~1+\left(\dfrac{3}{5}\right)^x+\dfrac{1}{x\cdot {5^x}}~~}$

$L=\dfrac{1-0+0}{~~1+0+0~~}\\\\ L=1\\\\ \boxed{\lim_{x\to\infty}\dfrac{5^x-3^x+1}{5^x+3^x+\frac{1}{x}}=1}$