\( \begin{array}{lll}\text { 30. } \lim \frac{e^{2}}{x^{2}}= & \text { B. } 0 & \text { C. } 1 \\ \begin{array}{lll}\text { A. }-e & \text { D. e } & \text { E. The limit does not e }\end{array} \\ \text { alculator Free Response } & & \end{array} \)

To evaluate the limit: \[ \lim_{x \to \infty} \frac{e^2}{x^2} \] Let's analyze the behavior of the function as \( x \) approaches infinity. 1. **Numerator (\( e^2 \))**: This is a constant value (\( e \) is Euler's number, approximately 2.71828), so as \( x \) changes, the numerator remains \( e^2 \). 2. **Denominator (\( x^2 \))**: As \( x \) approaches infinity, \( x^2 \) grows without bound (i.e., it becomes very large). Therefore, as \( x \) becomes larger and larger, the fraction \( \frac{e^2}{x^2} \) becomes smaller and smaller because the denominator increases much faster than the numerator. \[ \lim_{x \to \infty} \frac{e^2}{x^2} = 0 \] **Answer:** **B. 0**