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

To evaluate the limit \(\lim_{x \to \infty} \left(\frac{2x-3}{2x+1}\right)^{5x}\), we can use the following steps:<br /><br />1. **Simplify the expression inside the limit:**<br /><br /> \[<br /> \frac{2x-3}{2x+1} = \frac{2 - \frac{3}{x}}{2 + \frac{1}{x}}<br /> \]<br /><br /> As \(x \to \infty\), the terms \(\frac{3}{x}\) and \(\frac{1}{x}\) approach 0. Therefore, the expression simplifies to:<br /><br /> \[<br /> \frac{2 - \frac{3}{x}}{2 + \frac{1}{x}} \approx \frac{2}{2} = 1<br /> \]<br /><br />2. **Rewrite the limit in terms of the simplified expression:**<br /><br /> \[<br /> \lim_{x \to \infty} \left(\frac{2x-3}{2x+1}\right)^{5x} = \lim_{x \to \infty} \left(1 - \frac{4}{2x+1}\right)^{5x}<br /> \]<br /><br />3. **Use the approximation \(1 - \frac{4}{2x+1} \approx \frac{4}{2x+1}\) as \(x \to \infty\):**<br /><br /> \[<br /> \lim_{x \to \infty} \left(1 - \frac{4}{2x+1}\right)^{5x} \approx \lim_{x \to \infty} \left(\frac{4}{2x+1}\right)^{5x}<br /> \]<br /><br />4. **Simplify the exponent:**<br /><br /> \[<br /> \left(\frac{4}{2x+1}\right)^{5x} = \left(\frac{4}{2x+1}\right)^{5x} = \left(4^{5x}\right) \left((2x+1)^{-5x}\right)<br /> \]<br /><br />5. **Evaluate the limit of each part separately:**<br /><br /> - The term \(4^{5x}\) grows exponentially as \(x \to \infty\).<br /> - The term \((2x+1)^{-5x}\) can be analyzed using Stirling's approximation or the properties of exponential functions.<br /><br /> Specifically, \((2x+1)^{-5x}\) behaves like \((2x)^{-5x}\) for large \(x\):<br /><br /> \[<br /> (2x)^{-5x} = \left(2x\right)^{-5x} = \left(e^{\ln(2x)}\right)^{-5x} = e^{-5x \ln(2x)}<br /> \]<br /><br /> As \(x \to \infty\), \(\ln(2x) \approx x \ln(2)\), so:<br /><br /> \[<br /> -5x \ln(2x) \approx -5x \cdot x \ln(2) = -5x^2 \ln(2)<br /> \]<br /><br /> Therefore,<br /><br /> \[<br /> e^{-5x \ln(2x)} \to e^{-\infty} = 0<br /> \]<br /><br />6. **Combine the results:**<br /><br /> Since \(4^{5x}\) grows exponentially and \((2x+1)^{-5x}\) decays exponentially, the overall expression tends to:<br /><br /> \[<br /> \lim_{x \to \infty} \left(\frac{4}{2x+1}\right)^{5x} = \infty<br /> \]<br /><br />Thus, the limit is:<br /><br />\[<br />\lim_{x \to \infty} \left(\frac{2x-3}{2x+1}\right)^{5x} = \infty<br />\]