\( \left. \begin{array} { l } { f ( x ) = \left\{ \begin{array} { l } { x ^ { 2 } , x \leq 1 } \\ { 1 - x , x > 1 } \end{array} \right.} \\ { \lim _ { x \rightarrow 1 + } f ( x ) = } \end{array} \right. \)

To find the limit of the function \( f(x) \) as \( x \) approaches \( 1 \) from the right (denoted as \( \lim_{x \rightarrow 1^+} f(x) \)), we need to consider the definition of the function for values of \( x \) greater than \( 1 \). The function \( f(x) \) is defined as follows: \[ f(x) = \begin{cases} x^2 & \text{if } x \leq 1 \\ 1 - x & \text{if } x > 1 \end{cases} \] Since we are interested in the limit as \( x \) approaches \( 1 \) from the right, we will use the second case of the function, which applies when \( x > 1 \): \[ f(x) = 1 - x \quad \text{for } x > 1 \] Now, we can compute the limit: \[ \lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1^+} (1 - x) \] As \( x \) approaches \( 1 \) from the right, \( 1 - x \) approaches \( 1 - 1 = 0 \). Thus, we have: \[ \lim_{x \rightarrow 1^+} f(x) = 0 \] Therefore, the final answer is: \[ \lim_{x \rightarrow 1^+} f(x) = 0 \]