2. If \( \lim _{n \rightarrow 2^{+}} f(x)=3 \quad \& \lim _{n \rightarrow 2^{-}} f(x)=5 \) then \( \lim _{n \rightarrow 2} f(x)=\cdots \) \( \begin{array}{ll}\text { (A) } 3 & \text { (B) } 5 \\ \text { (C) } 8 & \text { (D ) D N E } \\ \text { A } \\ \text { B } \\ \text { с }\end{array} \)

To determine \( \lim _{n \rightarrow 2} f(x) \), we need to analyze the given one-sided limits: 1. \( \lim _{n \rightarrow 2^{+}} f(x) = 3 \): This means that as \( n \) approaches 2 from the right (values greater than 2), the function \( f(x) \) approaches 3. 2. \( \lim _{n \rightarrow 2^{-}} f(x) = 5 \): This means that as \( n \) approaches 2 from the left (values less than 2), the function \( f(x) \) approaches 5. For the limit \( \lim _{n \rightarrow 2} f(x) \) to exist, both one-sided limits must be equal. In this case, we have: - From the right: \( 3 \) - From the left: \( 5 \) Since \( 3 \neq 5 \), the two one-sided limits are not equal. Therefore, the limit \( \lim _{n \rightarrow 2} f(x) \) does not exist. Thus, the answer is \( \text{(D) D N E} \).