Question

The value of \[ \lim_{x \to 1} \frac{\frac{1}{2x + 1} - \frac{1}{3}}{x - 1} \] is equal to:

• A. \( \frac{-2}{9} \)
• B. \( \frac{2}{9} \)
• C. \( \frac{-2}{3} \)
• D. \( \frac{2}{3} \)
• E. 0

Hint:

When evaluating limits involving fractions, first simplify the expression and look for common terms that can be canceled. If needed, use standard limit rules like L'Hopital's rule or direct substitution.

Answer 1

The Correct Option is A

We are asked to evaluate the limit: \[ \lim_{x \to 1} \frac{\frac{1}{2x + 1} - \frac{1}{3}}{x - 1} \] Step 1: Simplify the numerator
First, simplify the expression inside the numerator: \[ \frac{1}{2x + 1} - \frac{1}{3} \] To combine these fractions, find the common denominator: \[ \frac{1}{2x + 1} - \frac{1}{3} = \frac{3 - (2x + 1)}{3(2x + 1)} = \frac{3 - 2x - 1}{3(2x + 1)} = \frac{2 - 2x}{3(2x + 1)} \] Thus, the original expression becomes: \[ \frac{\frac{2 - 2x}{3(2x + 1)}}{x - 1} \] This simplifies to: \[ \frac{2(1 - x)}{3(2x + 1)(x - 1)} = \frac{-2(x - 1)}{3(2x + 1)(x - 1)} \] Step 2: Cancel out common factors
We can cancel out \( (x - 1) \) in the numerator and denominator: \[ \frac{-2}{3(2x + 1)} \] Step 3: Evaluate the limit
Now, substitute \( x = 1 \) into the simplified expression: \[ \lim_{x \to 1} \frac{-2}{3(2x + 1)} = \frac{-2}{3(2(1) + 1)} = \frac{-2}{3(3)} = \frac{-2}{9} \] Thus, the value of the limit is: \[ \frac{-2}{9} \] Thus, the correct answer is option (A), \( \frac{-2}{9} \).