Evaluate the limit: lim (x->0) [tan(2x) - sin(2x)] / x3

Step 1: Evaluate the limit by direct substitution. Substituting x=0 into the expression gives: ((2 · 0) - (2 · 0))/(0^3) = ((0) - (0))/(0) = (0 - 0)/(0) = (0)/(0) This is an indeterminate form, so we can use L'Hôpital's Rule or Taylor series expansions. We will use Taylor series expansions. Step 2: Write down the Taylor series expansions for (u) and (u) around u=0. The Taylor series for (u) up to the third order is: (u) = u + (u^3)/(3) + O(u^5) The Taylor series for (u) up to the third order is: (u) = u - (u^3)/(3!) + O(u^5) Step 3: Substitute u=2x into the Taylor series expansions. For (2x): (2x) = (2x) + ((2x)^3)/(3) + O((2x)^5) = 2x + (8x^3)/(3) + O(x^5) For (2x): (2x) = (2x) - ((2x)^3)/(3!) + O((2x)^5) = 2x - (8x^3)/(6) + O(x^5) = 2x - (4x^3)/(3) + O(x^5) Step 4: Substitute these expansions into the original limit expression. _x 0 ((2x + 8x^3)/(3) + O(x^5)) - (2x - (4x^3)/(3) + O(x^5))x^3 Step 5: Simplify the numerator. _x 0 (2x + 8x^3)/(3) - 2x + (4x^3)/(3) + O(x^5)x^3 Combine the x^3 terms: _x 0 ((8)/(3) + (4)/(3))x^3 + O(x^5)x^3 _x 0 (12)/(3)x^3 + O(x^5)x^3 _x 0 (4x^3 + O(x^5))/(x^3) Step 6: Divide each term in the numerator by x^3 and evaluate the limit. _x 0 ( (4x^3)/(x^3) + (O(x^5))/(x^3) ) _x 0 ( 4 + O(x^2) ) As x 0, the term O(x^2) approaches 0. Therefore, the limit is: 4 + 0 = 4 The final answer is 4. Send me the next one 📸