Questions: Use l'Hospital's Rule to evaluate the limit. [ lim x rightarrow-4 frac-3 x-123 x^2+18 x+24= ]

Solution Steps

To evaluate the limit using l'Hospital's Rule, we first check if the limit is in an indeterminate form like \(\frac{0}{0}\). If it is, we differentiate the numerator and the denominator separately and then evaluate the limit again. Repeat the process if necessary until the limit is no longer in an indeterminate form.

First, we need to determine if the limit is in an indeterminate form. We substitute \( x = -4 \) into the numerator and the denominator:

• Numerator: \(-3(-4) - 12 = 12 - 12 = 0\)
• Denominator: \(3(-4)^2 + 18(-4) + 24 = 3(16) - 72 + 24 = 48 - 72 + 24 = 0\)

Since both the numerator and the denominator evaluate to 0, the limit is in the indeterminate form \(\frac{0}{0}\).

Since the limit is in the form \(\frac{0}{0}\), we can apply l'Hospital's Rule, which states that:

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]

provided the limit on the right exists. We need to find the derivatives of the numerator and the denominator.

• Derivative of the numerator, \(-3x - 12\), is \(-3\).
• Derivative of the denominator, \(3x^2 + 18x + 24\), is \(6x + 18\).

Now, we evaluate the limit of the derivatives:

\[ \lim_{x \to -4} \frac{-3}{6x + 18} \]

Substitute \( x = -4 \) into the derivative of the denominator:

\[ 6(-4) + 18 = -24 + 18 = -6 \]

Thus, the limit becomes:

\[ \lim_{x \to -4} \frac{-3}{-6} = \frac{1}{2} \]

Final Answer