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

To find the limit of the given expression as \( x \) approaches infinity, we can use the following steps:<br /><br />1. Divide both the numerator and denominator by the highest power of \( x \) in the denominator.<br />2. Simplify the expression.<br />3. Evaluate the limit.<br /><br />Let's apply these steps to the given expression:<br /><br />\[<br />\lim_{x \to \infty} \frac{2x^2 + 6x - 5}{5x^2 - x - 1}<br />\]<br /><br />Step 1: Divide both the numerator and denominator by \( x^2 \):<br /><br />\[<br />\lim_{x \to \infty} \frac{\frac{2x^2 + 6x - 5}{x^2}}{\frac{5x^2 - x - 1}{x^2}} = \lim_{x \to \infty} \frac{2 + \frac{6}{x} - \frac{5}{x^2}}{5 - \frac{1}{x} - \frac{1}{x^2}}<br />\]<br /><br />Step 2: Simplify the expression:<br /><br />\[<br />\lim_{x \to \infty} \frac{2 + \frac{6}{x} - \frac{5}{x^2}}{5 - \frac{1}{x} - \frac{1}{x^2}}<br />\]<br /><br />As \( x \) approaches infinity, the terms \( \frac{6}{x} \), \( \frac{5}{x^2} \), \( \frac{1}{x} \), and \( \frac{1}{x^2} \) all approach 0. Therefore, the expression simplifies to:<br /><br />\[<br />\frac{2}{5}<br />\]<br /><br />Step 3: Evaluate the limit:<br /><br />\[<br />\lim_{x \to \infty} \frac{2 + \frac{6}{x} - \frac{5}{x^2}}{5 - \frac{1}{x} - \frac{1}{x^2}} = \frac{2}{5}<br />\]<br /><br />Therefore, the limit of the given expression as \( x \) approaches infinity is \( \frac{2}{5} \).