Step 6 1- - cos(x) After applying L'Hospital's Rule twice, we have lim X-0 48x2 The derivative of 1 cos(x) with respect to x is sin(x) The derivative of 48x2 with respect to x is 96x ✓ 96x Step 7 Since the derivative of 1 - cos(x) is sin(x) and the derivative of 48x² is 96x, sin(x) 1 - cos(x) lim X-0 48x² = lim x-0 96x Analyzing this we see that as x→ 0, sin(x) → 0 and 9 0 Step 8 After applying L'Hospital's Rule three times, we have lim So, we still 1 The derivative of sin(x) with respect to x is 96 The derivative of 96x with respect to x is 1 96 sin(x) x-0 96x X . x So, we still sin(x 1- cos(x) So, we still have an indeterminate limit of type T We will apply L'Hos lim X→0 48x² s sin(x) sin(x) 96x the derivative of 48x² is 96x, applying L'Hospital's Rule a third time gives us the follow 0 and 96x → 0 0 sin(x) ve have lim . So, we still have an indeterminate limit of type. We will apply L'H 1 96 6 x-0 96x X bly L'Hospital's Rule for a third time. To do so, we need to find additional derivatives. the following. I apply Hospital's Rule for a fourth time. To do so, we need to find additional derivatives.