Final answer:
L'Hôpital's Rule does not apply to B: Lim₁→−₉ (x²−81/x-−9) because it does not create an indeterminate form, unlike the other options where the rule could potentially be applied.
Step-by-step explanation:
L'Hôpital's Rule applies to indeterminate forms like 0/0 or ∞/∞. Let's evaluate each limit.
- Lim₁→₀ x²cos(x)/9x: As x approaches 0, the numerator and the denominator both approach 0. L'Hôpital's Rule can be applied.
- Lim₁→−₉ (x²−81/x-−9): The numerator (x² - 81) approaches 64 and the denominator (x + 9) approaches 0. This is not an indeterminate form and hence L'Hôpital's Rule does not apply.
- Lim₁→₁ x³−1/x-1: The numerator and the denominator both approach 0. L'Hôpital's Rule can be applied.
- Lim₁→₀ ln(x+1) −xsin(x): As x approaches 0, the expression ln(x+1) approaches ln (1) which is 0, and xsin(x) also approaches 0; however, since this condition does not yield an indeterminate form of the type 0/0 or ∞/∞, it is unclear without further analysis if L'Hôpital's Rule applies directly.
Therefore, L'Hôpital's Rule does not apply to option B: Lim₁→−₉ (x²−81/x-−9) because it is not an indeterminate form.