D) \( \lim _{x \rightarrow+\infty} \frac{2 x-3 x^{2}}{5 x^{2}+7} \)

To find the limit as \( x \) approaches infinity, we can divide the numerator and the denominator by \( x^2 \), which is the highest power of \( x \) in the denominator: \[ \lim_{x \rightarrow +\infty} \frac{2x - 3x^2}{5x^2 + 7} = \lim_{x \rightarrow +\infty} \frac{\frac{2}{x} - 3}{5 + \frac{7}{x^2}} \] As \( x \) approaches infinity, \( \frac{2}{x} \) approaches \( 0 \) and \( \frac{7}{x^2} \) approaches \( 0 \). Therefore, we have: \[ \lim_{x \rightarrow +\infty} \frac{0 - 3}{5 + 0} = \frac{-3}{5} \] Thus, the limit is: \[ \lim_{x \rightarrow +\infty} \frac{2x - 3x^2}{5x^2 + 7} = -\frac{3}{5} \]