1. lim _(n arrow infty)(sqrt(n^2)+5 n+1-sqrt(n^2)+3 n-1)

To find the limit of the given expression as \( n \) approaches infinity, we can use the technique of rationalizing the numerator. Here's the step-by-step solution:<br /><br />Given expression:<br />\[ \lim_{n \rightarrow \infty} \left( \sqrt{n^2 + 5n + 1} - \sqrt{n^2 + 3n - 1} \right) \]<br /><br />First, we multiply and divide the expression by the conjugate of the numerator:<br />\[ \lim_{n \rightarrow \infty} \frac{\left( \sqrt{n^2 + 5n + 1} - \sqrt{n^2 + 3n - 1} \right) \cdot \left( \sqrt{n^2 + 5n + 1} + \sqrt{n^2 + 3n - 1} \right)}{\sqrt{n^2 + 5n + 1} + \sqrt{n^2 + 3n - 1}} \]<br /><br />The numerator simplifies as follows:<br />\[ \left( \sqrt{n^2 + 5n + 1} - \sqrt{n^2 + 3n - 1} \right) \cdot \left( \sqrt{n^2 + 5n + 1} + \sqrt{n^2 + 3n - 1} \right) = (n^2 + 5n + 1) - (n^2 + 3n - 1) = 2n + 2 \]<br /><br />So the expression becomes:<br />\[ \lim_{n \rightarrow \infty} \frac{2n + 2}{\sqrt{n^2 + 5n + 1} + \sqrt{n^2 + 3n - 1}} \]<br /><br />Next, we factor out \( n^2 \) from the square roots in the denominator:<br />\[ \sqrt{n^2 + 5n + 1} = n \sqrt{1 + \frac{5}{n} + \frac{1}{n^2}} \]<br />\[ \sqrt{n^2 + 3n - 1} = n \sqrt{1 + \frac{3}{n} - \frac{1}{n^2}} \]<br /><br />As \( n \) approaches infinity, the terms \( \frac{5}{n} \) and \( \frac{1}{n^2} \) in the first square root, and \( \frac{3}{n} \) and \( \frac{-1}{n^2} \) in the second square root, approach zero. Therefore, the denominator simplifies to:<br />\[ \sqrt{n^2 + 5n + 1} + \sqrt{n^2 + 3n - 1} \approx n + n = 2n \]<br /><br />So the expression simplifies to:<br />\[ \lim_{n \rightarrow \infty} \frac{2n + 2}{2n} = \lim_{n \rightarrow \infty} \frac{2n + 2}{2n} = \lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right) = 1 \]<br /><br />Therefore, the limit is:<br />\[ \boxed{1} \]