lim _(xarrow infty )((x^2+2x)/(x^2)+4)^(x-3)/(2)

To evaluate the limit \(\lim_{x \to \infty} \left( \frac{x^2 + 2x}{x^2 + 4} \right)^{\frac{x-), we can follow these steps:<br /><br />1. **Simplify the fraction inside the limit:**<br /><br /> \[<br /> \frac{x^2 + 2x}{x^2 + 4} = \frac{x^2(1 + \frac{2}{x})}{x^2(1 + \frac{4}{x^2})} = \frac{1 + \frac{2}{x}}{1 + \frac{4}{x^2}}<br /> \]<br /><br /> As \(x \to \infty\), \(\frac{2}{x} \to 0\) and \(\frac{4}{x^2} \to 0\). So,<br /><br /> \[<br /> \frac{1 + \frac{2}{x}}frac{4}{x^2}} \to \frac{1 + 0}{1 + 0} = 1<br /> \]<br /><br />2. **Rewrite the limit expression:**<br /><br /> \[<br /> \lim_{x \to \infty} \left( \frac{x^2 + 2x}{x^2 + 4} \right)^{\frac{x-3}{2}} = \lim_{x \to \infty} \left( 1 \right)^{\frac{x-3}{2}}<br /> \]<br /><br />3. **Evaluate the exponent:**<br /><br /> Since \(1^k = 1\) for any real number \(k\),<br /><br /> \[<br /> \left( 1 \right)^{\frac{x-3}{2}} = 1<br /> \]<br /><br />Therefore:<br /><br />\[<br />\lim_{x \to \infty} \left( \frac{x^2 + 2x}{x^2 + 4} \right)^{\frac{x-3}{2}} = 1<br />\]