Question

The value of \( \lim_{x \to 0} \frac{4x^3 - 2x^2 + x}{3x^2 + 2x} \) is \(\underline{\hspace{2cm}}\) (correct up to one decimal place).

Hint:

When you encounter an indeterminate form like \( \frac{0}{0} \), apply L'Hôpital's Rule to find the limit.

Answer 1

Correct Answer: 0.5

We need to find the limit: \[ \lim_{x \to 0} \frac{4x^3 - 2x^2 + x}{3x^2 + 2x}. \] Substituting \( x = 0 \) into the expression directly gives an indeterminate form \( \frac{0}{0} \). Hence, we use L'Hôpital's Rule: differentiate the numerator and denominator separately. The numerator's derivative is: \[ \frac{d}{dx}(4x^3 - 2x^2 + x) = 12x^2 - 4x + 1. \] The denominator's derivative is: \[ \frac{d}{dx}(3x^2 + 2x) = 6x + 2. \] Now, substitute \( x = 0 \): \[ \lim_{x \to 0} \frac{12x^2 - 4x + 1}{6x + 2} = \frac{1}{2}. \] Thus, the value of the limit is \( 0.5 \).