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

To evaluate the limit \[ \lim_{x \rightarrow +\infty} \frac{e^x}{2 + \cos(x)}, \] 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. Specifically, \( e^x \rightarrow +\infty \) as \( x \rightarrow +\infty \). 2. **Denominator**: The term \( 2 + \cos(x) \) oscillates between \( 1 \) and \( 3 \) because \( \cos(x) \) oscillates between \( -1 \) and \( 1 \). Therefore, \( 2 + \cos(x) \) will always be bounded and will not approach infinity. Specifically, it will oscillate between \( 1 \) and \( 3 \). Now, we can analyze the limit: \[ \frac{e^x}{2 + \cos(x)}. \] As \( x \) approaches infinity, the numerator \( e^x \) grows without bound, while the denominator \( 2 + \cos(x) \) remains bounded between \( 1 \) and \( 3 \). Thus, we can conclude that: \[ \lim_{x \rightarrow +\infty} \frac{e^x}{2 + \cos(x)} = \frac{+\infty}{\text{bounded}} = +\infty. \] Therefore, the limit is \[ \boxed{+\infty}. \]