\( x \rightarrow 7 \frac { \sqrt { x + 2 } - 3 } { x - 7 } = \)

To solve the limit \( \lim_{x \rightarrow 7} \frac { \sqrt { x + 2 } - 3 } { x - 7 } \), we can start by substituting \( x = 7 \) directly into the expression. This gives us \( \frac { \sqrt { 7 + 2 } - 3 } { 7 - 7 } = \frac { \sqrt{9} - 3 } { 0} = \frac {3 - 3} {0} = \frac {0} {0} \), which is indeterminate. To resolve this, we can use algebraic manipulation by multiplying the numerator and denominator by the conjugate of the numerator: \[ \frac { \sqrt { x + 2 } - 3 } { x - 7 } \cdot \frac { \sqrt { x + 2 } + 3 } { \sqrt { x + 2 } + 3 } = \frac { (x + 2) - 9 } { (x - 7)(\sqrt { x + 2 } + 3) } = \frac { x - 7 } { (x - 7)(\sqrt { x + 2 } + 3) } \] Now we can cancel out the \( x - 7 \) term (for \( x \neq 7 \)): \[ = \frac { 1 } { \sqrt { x + 2 } + 3 } \] Next, we can substitute \( x = 7 \) again to find the limit: \[ = \frac { 1 } { \sqrt { 7 + 2 } + 3 } = \frac { 1 } { 3 + 3 } = \frac { 1 } { 6 } \] Thus, the final answer is: \[ \lim_{x \rightarrow 7} \frac { \sqrt { x + 2 } - 3 } { x - 7 } = \frac{1}{6} \]