Applying L’Hôpital’s Rule to Trigonometric Limits

First, substitute x=0. Both numerator and denominator are sin(0)=0, so we have 0/0. Apply L'Hôpital's Rule. Let f(x) = sin(7x) and g(x) = sin(3x). Then f'(x) = 7cos(7x) and g'(x) = 3cos(3x). The new limit is limit as x approaches 0 of (7cos(7x)) / (3cos(3x)). Now we can evaluate by direct substitution: 7cos(0) / (3cos(0)) = 7 * 1 / (3 * 1) = 7/3. Therefore, the original limit is 7/3.

Substitute x=0: numerator is 1 - cos(0) = 1 - 1 = 0, denominator is 0. So it's 0/0. Apply L'Hôpital's Rule. Let f(x) = 1 - cos(2x) and g(x) = x. Then f'(x) = 2sin(2x) and g'(x) = 1. The new limit is limit as x approaches 0 of 2sin(2x) / 1 = 2sin(0) = 0. Therefore, the original limit is 0.

L'Hôpital's Rule directly applies to limits of the form 0/0 or infinity/infinity. However, other indeterminate forms like 0 * infinity, infinity - infinity, 0^0, 1^infinity, and infinity^0 can often be manipulated (e.g., by rewriting as a fraction) into a 0/0 or infinity/infinity form so that L'Hôpital's Rule can be used indirectly. Among the given options, 0 * infinity is an indeterminate form that can be converted into a ratio for applying L'Hôpital's Rule. Options A and B are not indeterminate; they tend to infinity (or negative infinity depending on signs). Option D is just 0.

As x approaches infinity, both numerator and denominator approach infinity. So we have infinity/infinity. Apply L'Hôpital's Rule. Let f(x) = 4x³ + 2x and g(x) = 7x³ - 5x². Then f'(x) = 12x² + 2 and g'(x) = 21x² - 10x. The new limit is limit as x approaches infinity of (12x² + 2) / (21x² - 10x). This is still infinity/infinity. Apply L'Hôpital's Rule again: f''(x) = 24x and g''(x) = 42x - 10. Limit becomes limit as x approaches infinity of (24x) / (42x - 10), still infinity/infinity. Apply L'Hôpital's Rule a third time: f'''(x) = 24 and g'''(x) = 42. The limit is now 24/42 = 4/7. Therefore, the original limit is 4/7.

Substitute x=0: numerator is e^0 - 0 - 1 = 1 - 1 = 0, denominator is 0. So it's 0/0. Apply L'Hôpital's Rule. Let f(x) = e^(4x) - 4x - 1, so f'(x) = 4e^(4x) - 4. Let g(x) = x², so g'(x) = 2x. The new limit is limit as x approaches 0 of (4e^(4x) - 4) / (2x) = limit as x approaches 0 of (4(e^(4x) - 1)) / (2x) = limit as x approaches 0 of (2(e^(4x) - 1)) / x. This is still 0/0 (since e^(0)-1=0). Apply L'Hôpital's Rule again. For the numerator, derivative of 2(e^(4x) - 1) is 2 * 4e^(4x) = 8e^(4x). The denominator's derivative is 1. So the limit becomes limit as x approaches 0 of 8e^(4x) / 1 = 8e^0 = 8. Therefore, the original limit is 8.

L'Hôpital's Rule can be applied repeatedly as long as the resulting limit after differentiation continues to be an indeterminate form of type 0/0 or infinity/infinity. Therefore, if after one application we still have 0/0, we can apply the rule again, provided the new functions are differentiable. We continue until we reach a determinate form or find that the limit does not exist.

First, substitute x = π/2. cos(π/2) = 0, and the denominator is pi/2 - pi/2 = 0. So we have 0/0. Apply L'Hôpital's Rule. Let f(x) = cos(x) and g(x) = x - π/2. Then f'(x) = -sin(x) and g'(x) = 1. The new limit is the limit as x approaches π/2 of -sin(x) / 1 = -sin(π/2) = -1. Therefore, the original limit is -1.

Substitute x=0: numerator is 5^0 - 1 = 1 - 1 = 0, denominator is 0. So it's 0/0. Apply L'Hôpital's Rule. Let f(x) = 5^x - 1 and g(x) = x. The derivative of 5^x is 5^x * ln(5) (using the rule for exponential functions). So f'(x) = 5^x * ln(5). g'(x) = 1. The new limit is limit as x approaches 0 of (5^x * ln(5)) / 1 = 5^0 * ln(5) = 1 * ln(5) = ln(5). Therefore, the original limit is ln(5).

The limit is of the form 0 * (-∞) as x approaches 0 from the right (since ln(x) approaches -∞). This is an indeterminate form, but not directly a ratio. To use L'Hôpital's Rule, we rewrite it as a ratio. One way is to write x * ln(x) = ln(x) / (1/x). Now as x approaches 0+, ln(x) approaches -∞ and 1/x approaches infinity, so we have the form -∞/infinity, which is an indeterminate form suitable for L'Hôpital's Rule. Therefore, option C is correct. Direct substitution (A) gives an undefined expression. Option B is incorrect because L'Hôpital's Rule applies to ratios, not products directly. Option D is false because the limit exists (it is 0).

Substitute x=1: numerator is 1 - 1 = 0, denominator is ln(1) = 0. So it's 0/0. Apply L'Hôpital's Rule. Let f(x) = x² - 1, so f'(x) = 2x. Let g(x) = ln(x), so g'(x) = 1/x. The new limit is limit as x approaches 1 of (2x) / (1/x) = limit as x approaches 1 of 2x². Now evaluate by direct substitution: 2 * 1² = 2. Therefore, the original limit is 2.

L'Hôpital's Rule is specifically for limits of the form f(x)/g(x) where both f and g tend to 0 or both tend to infinity (or negative infinity). For the form 0 * infinity, we must first rewrite the expression as a ratio (e.g., 0 * infinity = 0/(1/infinity) = 0/0 or infinity/(1/0)=infinity/infinity) before applying L'Hôpital's Rule. So it cannot be applied directly to the product; it must be converted to a ratio first. Therefore, statement C is false as written because it says "applied if the limit is of the form 0 * infinity" without mentioning the need for conversion. The other statements are true.

Substitute x=0: arctan(0)=0, denominator=0, so 0/0. Apply L'Hôpital's Rule. Let f(x) = arctan(3x), so f'(x) = 1/(1+(3x)²) * 3 = 3/(1+9x²). Let g(x)=x, so g'(x)=1. The new limit is limit as x approaches 0 of (3/(1+9x²)) / 1 = 3/(1+0) = 3. Therefore, the original limit is 3.

L'Hôpital's Rule states that if the limit of f'(x)/g'(x) exists (or is infinite), then it equals the original limit. If the limit of f'(x)/g'(x) does not exist (e.g., it oscillates), then L'Hôpital's Rule does not provide information about the original limit; the original limit might still exist or might not. In such cases, we need to use other methods to evaluate the original limit. Therefore, we cannot conclude that the original limit does not exist solely because the derivative ratio oscillates.

As x approaches infinity, both numerator and denominator approach infinity. So we have infinity/infinity. Apply L'Hôpital's Rule. Let f(x) = x² + sin(x), so f'(x) = 2x + cos(x). Let g(x) = 3x² + 5, so g'(x) = 6x. The new limit is limit as x approaches infinity of (2x + cos(x)) / (6x). This is still infinity/infinity? Actually, we can simplify: divide numerator and denominator by x: (2 + cos(x)/x) / 6. As x approaches infinity, cos(x)/x approaches 0 because cos(x) is bounded between -1 and 1. So the limit becomes (2 + 0)/6 = 1/3. Alternatively, we could apply L'Hôpital's Rule again if necessary, but the simplification works. Therefore, the original limit is 1/3.

Substitute x=0: numerator is sin(0)-0=0, denominator=0, so 0/0. Apply L'Hôpital's Rule. Let f(x)= sin(2x)-2x, f'(x)=2cos(2x)-2. g(x)=x³, g'(x)=3x². New limit: limit as x approaches 0 of (2cos(2x)-2)/(3x²). At x=0, this is (2-2)/0=0/0. Apply L'Hôpital's Rule again. f2(x)=2cos(2x)-2, f2'(x)=-4sin(2x). g2(x)=3x², g2'(x)=6x. New limit is limx→0 (-4sin(2x))/(6x) = limx→0(-2/3) * (sin(2x)/x). Apply L'Hôpital's Rule once more to sin(2x)/x so that limx→0(2cos(2x))/1 = 2, then multiply by -2/3. Therefore, the original limit is -4/3.