1. The problem is to evaluate a limit where direct substitution results in an indeterminate form, and we are asked to apply L'Hospital's Rule three times. 2. L'Hospital's Rule states that if the limit of $\frac{f(x)}{g(x)}$ as $x$ approaches a value results in $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then the limit can be found by differentiating numerator and denominator separately: $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$ provided the latter limit exists. 3. Applying L'Hospital's Rule three times means differentiating numerator and denominator three times each, then evaluating the limit. 4. Without a specific function, the general approach is: - Compute $f'(x)$ and $g'(x)$, evaluate the limit. - If still indeterminate, compute $f''(x)$ and $g''(x)$, evaluate the limit. - If still indeterminate, compute $f'''(x)$ and $g'''(x)$, evaluate the limit. 5. After the third differentiation, if the limit is determinate, that is the value of the original limit. 6. This method is useful when the original limit is complicated and repeatedly applying L'Hospital's Rule simplifies the expression. Final answer: Apply L'Hospital's Rule up to three times by differentiating numerator and denominator each time until the limit is no longer indeterminate, then evaluate the limit.