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

To solve this limit, we can use the properties of exponents and the fact that the limit of a constant raised to infinity is 1.<br /><br />First, let's rewrite the expression inside the parentheses:<br /><br />$\frac{2x^2 + 3}{2x^2 - 4} = \frac{2x^2(1 + \frac{3}{2x^2})}{2x^2(1 - \frac{4}{2x^2})} = \frac{1 + \frac{3}{2x^2}}{1 - \frac{4}{2x^2}}$<br /><br />Now, we can rewrite the original expression as:<br /><br />$\left(\frac{1 + \frac{3}{2x^2}}{1 - \frac{4}{2x^2}}\right)^{3x}$<br /><br />Next, we can use the property of exponents that states $(1 + \frac{a}{x})^x \approx e^a$ as $x$ approaches infinity:<br /><br />$\left(\frac{1 + \frac{3}{2x^2}}{1 - \frac{4}{2x^2}}\right)^{3x} \approx \left(e^{\frac{3}{2x^2}}\right)^{3x} = e^{\frac{9}{2x^2}}$<br /><br />As $x$ approaches infinity, $\frac{9}{2x^2}$ approaches 0, so the expression simplifies to:<br /><br />$e^0 = 1$<br /><br />Therefore, the limit is $\boxed{1}$.