Question

The value of \( y \) for which the following limit exists is \[ \lim_{x \to 1} \frac{2x^2 - yx - x + 3}{3x^2 - 5x + 2} \]

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

- For limits that result in an indeterminate form like \( \frac{0}{0} \), check if both the numerator and denominator approach 0.
- Use L'Hopital's Rule or factor the expression to resolve the indeterminate form.

Answer 1

The Correct Option is C

For the limit to exist, the numerator and denominator must both approach 0 as \( x \to 1 \), so that we can apply L'Hopital's Rule or factor the expression. Substitute \( x = 1 \) into the denominator: \[ 3(1)^2 - 5(1) + 2 = 3 - 5 + 2 = 0 \] The denominator approaches 0 at \( x = 1 \). Therefore, the numerator must also approach 0 at \( x = 1 \) for the limit to exist. Substitute \( x = 1 \) into the numerator: \[ 2(1)^2 - y(1) - 1 + 3 = 2 - y - 1 + 3 = 4 - y \] For the numerator to approach 0, we set: \[ 4 - y = 0 \Rightarrow y = 4 \] Thus, the value of \( y \) for which the limit exists is \( \boxed{4} \).