$\lim _{x \rightarrow 5} \frac{x+2}{x^{2}-5}$

Step 1 ：We are given the function \(f(x) = \frac{x+2}{x^{2}-5}\) and we are asked to find the limit as x approaches 5.

Step 2 ：Substituting x=5 into the function, we get a division by zero, which is undefined. Therefore, we need to simplify the function first.

Step 3 ：The denominator is a difference of squares, which can be factored into \((x-\sqrt{5})(x+\sqrt{5})\). However, the numerator does not contain these factors, so we cannot simplify the function by factoring.

Step 4 ：Therefore, we need to use L'Hopital's rule, which states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives.

Step 5 ：Taking the derivative of the numerator, we get 1. Taking the derivative of the denominator, we get -2x.

Step 6 ：Substituting these into the function, we get \(f'(x) = \frac{1}{-2x}\).

Step 7 ：Substituting x=5 into this function, we get \(-\frac{1}{10}\).

Step 8 ：Final Answer: The limit of the function as x approaches 5 is \(\boxed{-\frac{1}{8}}\).