\( \lim _ { x \rightarrow + \infty } \frac { e ^ { x } } { 2 + n + x } = \)

To evaluate the limit \[ \lim_{x \rightarrow +\infty} \frac{e^x}{2 + n + x}, \] we start by analyzing the behavior of the numerator and the denominator as \( x \) approaches infinity. 1. **Numerator**: The term \( e^x \) grows exponentially as \( x \) increases. 2. **Denominator**: The term \( 2 + n + x \) grows linearly since \( x \) dominates the constant terms \( 2 \) and \( n \). As \( x \) approaches infinity, the exponential function \( e^x \) will grow much faster than the linear function \( 2 + n + x \). Therefore, we can expect the limit to approach infinity. To make this more rigorous, we can divide both the numerator and the denominator by \( e^x \): \[ \frac{e^x}{2 + n + x} = \frac{1}{\frac{2 + n + x}{e^x}}. \] Now, we need to analyze the limit of the denominator: \[ \lim_{x \rightarrow +\infty} \frac{2 + n + x}{e^x}. \] As \( x \) approaches infinity, the term \( x \) in the numerator will dominate the constants \( 2 \) and \( n \), so we can simplify this to: \[ \lim_{x \rightarrow +\infty} \frac{x}{e^x}. \] Using L'Hôpital's Rule, since both the numerator and denominator approach infinity, we differentiate the numerator and denominator: - The derivative of \( x \) is \( 1 \). - The derivative of \( e^x \) is \( e^x \). Applying L'Hôpital's Rule gives us: \[ \lim_{x \rightarrow +\infty} \frac{x}{e^x} = \lim_{x \rightarrow +\infty} \frac{1}{e^x} = 0. \] Thus, we have: \[ \lim_{x \rightarrow +\infty} \frac{2 + n + x}{e^x} = 0. \] Returning to our original limit, we find: \[ \lim_{x \rightarrow +\infty} \frac{e^x}{2 + n + x} = \frac{1}{0} = +\infty. \] Therefore, the final result is: \[ \lim_{x \rightarrow +\infty} \frac{e^x}{2 + n + x} = +\infty. \]