\( \lim _ { x \rightarrow 2 } \frac { \sigma \cup v x } { f ( x ) + 1 } \)

To evaluate the limit \[ \lim_{x \rightarrow 2} \frac{\sigma \cup v x}{f(x) + 1}, \] we need to clarify the terms involved in the expression. 1. **Understanding the notation**: - The notation \(\sigma \cup v x\) is not standard in calculus. It could represent a specific function or operation depending on the context. If \(\sigma\) and \(v\) are constants or functions, we need to know their definitions. - \(f(x)\) is presumably a function defined for values around \(x = 2\). 2. **Behavior of the limit**: - To compute the limit, we need to know the behavior of both the numerator and the denominator as \(x\) approaches 2. - If \(f(2) + 1 \neq 0\), then we can directly substitute \(x = 2\) into the limit. - If \(f(2) + 1 = 0\), we would need to analyze the limit further, possibly using L'Hôpital's rule if we encounter an indeterminate form. 3. **Substituting values**: - If we assume \(\sigma \cup v x\) simplifies to a specific value or function as \(x\) approaches 2, we can substitute that value into the limit. Without additional context or definitions for \(\sigma\), \(v\), and \(f(x)\), we cannot compute the limit explicitly. Please provide more information about these terms, and I would be happy to help you evaluate the limit!