This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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You're on a roll — here are the solutions for these limit problems. 1. Evaluate: _x 0 ( x)/(x) Step 1: This is a fundamental trigonometric limit. _x 0 ( x)/(x) = 1 The limit is 1. 2. Evaluate: _x 0 (1 - x)/(x^2) Step 1: This is a fundamental trigonometric limit. _x 0 (1 - x)/(x^2) = (1)/(2) The limit is (1)/(2). 3. Evaluate: _x ( x)/(x) Step 1: As x , both x and x , which is an indeterminate form ()/(). Apply L'Hôpital's Rule. _x ( x)/(x) = _x (d)/(dx)( x)(d)/(dx)(x) Step 2: Differentiate the numerator and the denominator. _x (1)/(x)1 = _x (1)/(x) Step 3: Evaluate the limit. _x (1)/(x) = 0 The limit is 0. 4. Evaluate: _x (x)/(e^x) Step 1: As x , both x and e^x , which is an indeterminate form ()/(). Apply L'Hôpital's Rule. _x (x)/(e^x) = _x (d)/(dx)(x)(d)/(dx)(e^x) Step 2: Differentiate the numerator and the denominator. _x (1)/(e^x) Step 3: Evaluate the limit. As x , e^x , so (1)/(e^x) 0. _x (1)/(e^x) = 0 The limit is 0. 5. Evaluate: _x 0 (e^x - 1)/(x) Step 1: This is a fundamental limit related to the exponential function. _x 0 (e^x - 1)/(x) = 1 The limit is 1. 6. Evaluate: _x 1 ( x)/(x - 1) Step 1: As x 1, x 1 = 0 and x - 1 1 - 1 = 0, which is an indeterminate form (0)/(0). Apply L'Hôpital's Rule. _x 1 ( x)/(x - 1) = _x 1 (d)/(dx)( x)(d)/(dx)(x - 1) Step 2: Differentiate the numerator and the denominator. _x 1 (1)/(x)1 = _x 1 (1)/(x) Step 3: Substitute x=1 into the expression. _x 1 (1)/(x) = (1)/(1) = 1 The limit is 1. 7. Evaluate: _x 0 ( x - x)/(x^3) Step 1: As x 0, x - x 0 - 0 = 0 and x^3 0, which is an indeterminate form (0)/(0). Apply L'Hôpital's Rule. _x 0 ( x - x)/(x^3) = _x 0 (^2 x - 1)/(3x^2) Step 2: This is still an indeterminate form (0)/(0) since ^2 0 - 1 = 1 - 1 = 0. Apply L'Hôpital's Rule again. _x 0 (^2 x - 1)/(3x^2) = _x 0 (2 x ( x x))/(6x) = _x 0 (2 ^2 x x)/(6x) = _x 0 (^2 x x)/(3x) Step 3: This is still an indeterminate form (0)/(0). Apply L'Hôpital's Rule a third time. _x 0 (^2 x x)/(3x) = _x 0 ((2 x ( x x)) x + ^2 x (^2 x))/(3) = _x 0 (2 ^2 x ^2 x + ^4 x)/(3) Step 4: Substitute x=0. (2 ^2 0 ^2 0 + ^4 0)/(3) = (2(1)^2(0)^2 + (1)^4)/(3) = (0 + 1)/(3) = (1)/(3) The limit is (1)/(3). 8. Evaluate: _x (x^2 + 1)/(e^x) Step 1: As x , both x^2 + 1 and e^x , which is an indeterminate form ()/(). Apply L'Hôpital's Rule. _x (x^2 + 1)/(e^x) = _x (d)/(dx)(x^2 + 1)(d)/(dx)(e^x) = _x (2x)/(e^x) Step 2: This is still an indeterminate form ()/(). Apply L'Hôpital's Rule again. _x (2x)/(e^x) = _x (d)/(dx)(2x)(d)/(dx)(e^x) = _x (2)/(e^x) Step 3: Evaluate the limit. As x , e^x , so (2)/(e^x) 0. _x (2)/(e^x) = 0 The limit is 0. What's next?