6.114 lim _(xarrow infty )((4x^2+2)/(4x^2)-1)^5x^(2)

To evaluate the limit \(\lim_{x \to \infty} \left( \frac{4x^2 + 2}{4x^2 - 1} \right)^{5x^2}\), we can use the following steps:<br /><br />1. **Rewrite the expression inside the limit:**<br /><br /> \[<br /> \frac{4x^2 + 2}{4x^2 - 1} = \frac{4x^2(1 + \frac{2}{4x^2})}{4x^2(1 - \frac{1}{4x^2})} = \frac{1 + \frac{2}{4x^2}}{1 - \frac{1}{4x^2}}<br /> \]<br /><br />2. **Simplify the fraction as \(x\) approaches infinity:**<br /><br /> \[<br /> \frac{1 + \frac{2}{4x^2}}{1 - \frac{1}{4x^2}} \approx 1 + \frac{2}{4x^2} \quad \text{and} \quad 1 - \frac{1}{4x^2} \approx 1<br /> \]<br /><br /> Therefore,<br /><br /> \[<br /> \frac{1 + \frac{2}{4x^2}}{1 - \frac{1}{4x^2}} \approx 1 + \frac{2}{4x^2} = 1 + \frac{1}{2x^2}<br /> \]<br /><br />3. **Substitute this approximation back into the original limit:**<br /><br /> \[<br /> \lim_{x \to \infty} \left( 1 + \frac{1}{2x^2} \right)^{5x^2}<br /> \]<br /><br />4. **Use the exponential approximation \( (1 + \frac{1}{n})^n \approx e \) as \( n \to \infty \):**<br /><br /> Let \( y = 5x^2 \). Then,<br /><br /> \[<br /> \left( 1 + \frac{1}{2x^2} \right)^{5x^2} = \left( 1 + \frac{1}{2x^2} \right)^{y/2}<br /> \]<br /><br /> As \( x \to \infty \), \( \frac{1}{2x^2} \to 0 \), so:<br /><br /> \[<br /> \left( 1 + \frac{1}{2x^2} \right)^{y/2} \approx e^{y/2}<br /> \]<br /><br />5. **Since \( y = 5x^2 \), we have:**<br /><br /> \[<br /> e^{y/2} = e^{5x^2/2}<br /> \]<br /><br />6. **Evaluate the limit:**<br /><br /> \[<br /> \lim_{x \to \infty} e^{5x^2/2} = \infty<br /> \]<br /><br />Therefore, the limit is:<br /><br />\[<br />\lim_{x \to \infty} \left( \frac{4x^2 + 2}{4x^2 - 1} \right)^{5x^2} = \infty<br />\]