6.117. lim _(xarrow infty )((5x^3-2)/(5x^3)+1)^-6x^(3)

To solve this limit, we can use the properties of exponents and the fact that as \( x \) approaches infinity, the dominant terms in the numerator and denominator will be the highest degree terms.<br /><br />First, let's rewrite the expression inside the limit:<br /><br />\[<br />\lim_{x \to \infty} \left( \frac{5x^3 - 2}{5x^3 + 1} \right)^{-6x^3}<br />\]<br /><br />We can simplify the fraction inside the limit:<br /><br />\[<br />\frac{5x^3 - 2}{5x^3 + 1} = \frac{5x^3 (1 - \frac{2}{5x^3})}{5x^3 (1 + \frac{1}{5x^3})} = \frac{1 - \frac{2}{5x^3}}{1 + \frac{1}{5x^3}}<br />\]<br /><br />As \( x \) approaches infinity, the terms \(\frac{2}{5x^3}\) and \(\frac{1}{5x^3}\) approach 0. Therefore, the fraction simplifies to:<br /><br />\[<br />\frac{1 - 0}{1 + 0} = 1<br />\]<br /><br />So, the expression inside the limit becomes:<br /><br />\[<br />\left( \frac{5x^3 - 2}{5x^3 + 1} \right)^{-6x^3} = 1^{-6x^3} = 1<br />\]<br /><br />Therefore, the limit is:<br /><br />\[<br />\lim_{x \to \infty} 1 = 1<br />\]<br /><br />Thus, the correct answer is:<br /><br />\[<br />\boxed{1}<br />\]