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

To solve this limit, we can use the properties of exponential and logarithmic functions. Let's rewrite the expression as:<br /><br />$\lim_{x\to\infty} \left(\frac{x^2+x-1}{x^2-2x+5}\right)^{-2x} = \lim_{x\to\infty} \left(\frac{x^2+x-1}{x^2-2x+5}\right)^{-2x}$<br /><br />Now, let's take the natural logarithm of both sides:<br /><br />$\ln\left(\lim_{x\to\infty} \left(\frac{x^2+x-1}{x^2-2x+5}\right)^{-2x}\right) = \ln\left(\lim_{x\to\infty} \left(\frac{x^2+x-1}{x^2-2x+5}\right)^{-2x}\right)$<br /><br />Using the property of logarithms, we can simplify the expression:<br /><br />$\lim_{x\to\infty} -2x \ln\left(\frac{x^2+x-1}{x^2-2x+5}\right)$<br /><br />Now, let's evaluate the limit:<br /><br />$\lim_{x\to\infty} -2x \ln\left(\frac{x^2+x-1}{x^2-2x+5}\right)$<br /><br />As $x$ approaches infinity, the terms $x^2$ in the numerator and denominator will dominate. So, we can simplify the expression further:<br /><br />$\lim_{x\to\infty} -2x \ln\left(1 + \frac{2x-4}{x^2-2x+5}\right)$<br /><br />As $x$ approaches infinity, the fraction $\frac{2x-4}{x^2-2x+5}$ approaches 0. Therefore, the natural logarithm term approaches 0:<br /><br />$\lim_{x\to\infty} -2x \cdot 0 = 0$<br /><br />So, the final answer is:<br /><br />$\lim_{x\to\infty} \left(\frac{x^2+x-1}{x^2-2x+5}\right)^{-2x} = e^0 = 1$