🔢 1. Calculus Q1. Evaluate the limit: lim ⁡ 𝑥 → 0 sin ⁡ 𝑥 − 𝑥 𝑥 3 x→0 lim ​ x 3 sinx−x ​

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We need to evaluate the limit of sine x minus x over x cubed as x approaches zero. When we substitute x equals zero directly, we get zero over zero, which is an indeterminate form. This means we need to use special techniques like L'Hôpital's rule to find the limit. Now we apply L'Hôpital's rule for the first time. We differentiate the numerator sine x minus x to get cosine x minus one, and differentiate the denominator x cubed to get three x squared. This gives us the limit of cosine x minus one over three x squared as x approaches zero. However, when we substitute x equals zero, we still get zero over zero, so we need to apply L'Hôpital's rule again. We apply L'Hôpital's rule for the second time. We differentiate the numerator cosine x minus one to get negative sine x, and differentiate the denominator three x squared to get six x. This gives us the limit of negative sine x over six x as x approaches zero. Unfortunately, this is still an indeterminate form zero over zero, so we need to apply L'Hôpital's rule one more time. Now we apply L'Hôpital's rule for the third and final time. We differentiate the numerator negative sine x to get negative cosine x, and differentiate the denominator six x to get six. This gives us the limit of negative cosine x over six as x approaches zero. Now we can directly substitute x equals zero to get negative cosine of zero over six, which equals negative one over six. This is our final answer. Let's summarize our complete solution. We started with the limit of sine x minus x over x cubed as x approaches zero. After applying L'Hôpital's rule three times, we transformed this into the limit of negative cosine x over six, which evaluates to negative one-sixth. Therefore, the final answer is negative one-sixth. This demonstrates how L'Hôpital's rule can be applied multiple times to resolve complex indeterminate forms.