Question

Evaluate the limit \[ \lim_{x \to 1} \frac{(\sqrt{x-1})(2x-3)}{2x^2 + x - 3} \]

• A. \( \frac{1}{10} \)
• B. 0
• C. 1
• D. \( -\frac{1}{10} \)

Hint:

When you encounter the indeterminate form \( \frac{0}{0} \), simplify the expression by factoring or using other algebraic techniques such as rationalizing the numerator or denominator.

Answer 1

The Correct Option is D

Step 1: Direct substitution.
Let's try substituting \( x = 1 \) directly into the given expression:
\[ \frac{(\sqrt{1-1})(2(1)-3)}{2(1)^2 + 1 - 3} = \frac{0 \cdot (-1)}{2 + 1 - 3} = \frac{0}{0} \]
We get an indeterminate form \( \frac{0}{0} \), so we need to simplify the expression.

Step 2: Simplify the numerator and denominator.
The numerator is \( (\sqrt{x-1})(2x-3) \). Factor the denominator \( 2x^2 + x - 3 \).
First, let's factor the quadratic \( 2x^2 + x - 3 \). We need two numbers that multiply to \( 2 \times -3 = -6 \) and add to \( 1 \).
These numbers are \( 3 \) and \( -2 \). So, we factor as: \[ 2x^2 + x - 3 = (2x - 3)(x + 1) \]

Step 3: Rewrite the expression.
Now, the expression becomes: \[ \frac{(\sqrt{x-1})(2x-3)}{(2x-3)(x+1)} \]
We can cancel \( (2x-3) \) from the numerator and denominator (as \( x \neq 1 \)) to simplify the expression: \[ \frac{\sqrt{x-1}}{x+1} \]

Step 4: Substitute \( x = 1 \).
Now, substitute \( x = 1 \) into the simplified expression: \[ \frac{\sqrt{1-1}}{1+1} = \frac{0}{2} = 0 \]

Step 5: Final result.
Thus, the limit is \( 0 \), corresponding to option (D).