Question Detail
Question
Evaluate the limit of \( \frac{x}{x+3} + \frac{x-3}{x^2-9} \) as \( x \) tends to 3.
Correct Answer
Option B is the correct answer.
Detailed Explanation
Simplify ( \frac{x-3}{x^2-9} = \frac{1}{x+3} ). The expression becomes ( \frac{x}{x+3} + \frac{1}{x+3} = \frac{x+1}{x+3} ). At ( x=3 ), this is ( \frac{4}{6} = \frac{2}{3} ).
Hint
Exercise 1.1, Question 6.