(i) Lt_(x rarr0)((tan 2x-sin 2x)/(x^(3)))

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(i)  Lt_(x rarr0)((tan 2x-sin 2x)/(x^(3)))

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Recognize Limit Problem: Recognize the limit problem and apply L'Hôpital's Rule if necessary. Since the limit is of the form 0/0, we can apply L'Hôpital's Rule, which states that if the limit of f(x)/g(x) as x approaches a value c is of the form 0/0 or ±∞/±∞, then the limit is the same as the limit of f'(x)/g'(x) as x approaches c, provided that this latter limit exists.Differentiate Numerator and Denominator: Differentiate the numerator and denominator separately. The derivative of the numerator tan(2x) - sin(2x) with respect to x is sec^2(2x)*2 - cos(2x)*2. The derivative of the denominator x^3 with respect to x is 3x^2.Apply L'Hôpital's Rule Once: Apply L'Hôpital's Rule once. We now have the new limit as x approaches 0 of (2*sec^2(2x) - 2*cos(2x)) / (3x^2).Evaluate New Limit: Evaluate the new limit as x approaches 0. As x approaches 0, sec^2(2x) approaches sec^2(0) which is 1, and cos(2x) approaches cos(0) which is also 1. So, the limit becomes (2*1 - 2*1) / (3*0^2) which simplifies to 0/0, indicating that we need to apply L'Hôpital's Rule again.Differentiate Again: Differentiate the new numerator and denominator again. The derivative of the new numerator 2*sec^2(2x) - 2*cos(2x) with respect to x is 2*2*sec^2(2x)*tan(2x)*2 - 2*(-sin(2x)*2). The derivative of the new denominator 3x^2 with respect to x is 6x.Apply L'Hôpital's Rule Second Time: Apply L'Hôpital's Rule a second time. We now have the new limit as x approaches 0 of (8*sec^2(2x)*tan(2x) - 4*sin(2x)) / 6x.Evaluate New Limit: Evaluate the new limit as x approaches 0. As x approaches 0, sec^2(2x) approaches 1, tan(2x) approaches 0, and sin(2x) approaches 0. So, the limit becomes (8*1*0 - 4*0) / 6*0 which simplifies to 0/0, indicating that we need to apply L'Hôpital's Rule yet again.Differentiate Again: Differentiate the new numerator and denominator again. The derivative of the new numerator 8*sec^2(2x)*tan(2x) - 4*sin(2x) with respect to x is 8*(2*sec^2(2x)*tan(2x)*2 + 2*sec^2(2x)*sec^2(2x)*2) - 4*(cos(2x)*2). The derivative of the new denominator 6x with respect to x is 6.Evaluate New Limit: Evaluate the new limit as x approaches 0. As x approaches 0, sec^2(2x) approaches 1, tan(2x) approaches 0, and cos(2x) approaches 1. So, the limit becomes (8*(2*1*0 + 2*1*1) - 4*(1*2)) / 6 which simplifies to (8*(0 + 2) - 8) / 6 which further simplifies to (16 - 8) / 6.Simplify Final Answer: Simplify the result to find the final answer. The final answer is (16 - 8) / 6 which simplifies to 8 / 6, and further simplifies to 4 / 3.

(i) $\newline$ $\lim_{x \to 0} \left( \frac{\tan 2x - \sin 2x}{x^{3}} \right)$

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(i) \newlinelim⁡x→0(tan⁡2x−sin⁡2xx3) \lim_{x \to 0} \left( \frac{\tan 2x - \sin 2x}{x^{3}} \right) limx→0​(x3tan2x−sin2x​)

(i) $\newline$ $\lim_{x \to 0} \left( \frac{\tan 2x - \sin 2x}{x^{3}} \right)$