L’Hôpital’s Rule: Deeper Concepts

First, we check the form of this limit. As x approaches 0, e^x - 1 approaches e^0 - 1 = 1 - 1 = 0, and x approaches 0, giving us a 0/0 indeterminate form. We can apply L'Hôpital's Rule by differentiating the numerator and denominator. The derivative of e^x - 1 is e^x, and the derivative of x is 1. So the limit becomes lim(x→0) e^x/1 = lim(x→0) e^x. As x approaches 0, e^x approaches e^0 = 1. Therefore, the limit equals 1.

First, we check the form of this limit. As x approaches 0, 1 - cos(x) approaches 1 - 1 = 0, and x² approaches 0, giving us a 0/0 indeterminate form. We can apply L'Hôpital's Rule by differentiating the numerator and denominator. The derivative of 1 - cos(x) is sin(x), and the derivative of x² is 2x. So the limit becomes lim(x→0) sin(x)/(2x). This is still a 0/0 form, so we apply L'Hôpital's Rule again. The derivative of sin(x) is cos(x), and the derivative of 2x is 2. So the limit becomes lim(x→0) cos(x)/2 = cos(0)/2 = 1/2.

First, we check the form of this limit. As x approaches infinity, x² approaches infinity and e^x approaches infinity, giving us an ∞/∞ indeterminate form. We can apply L'Hôpital's Rule by differentiating the numerator and denominator. The derivative of x² is 2x, and the derivative of e^x is e^x. So the limit becomes lim(x→∞) (2x)/(e^x). This is still an ∞/∞ form, so we apply L'Hôpital's Rule again. The derivative of 2x is 2, and the derivative of e^x is e^x. So the limit becomes lim(x→∞) 2/(e^x). As x approaches infinity, e^x grows without bound, so 2/(e^x) approaches 0. Therefore, the limit equals 0.

First, we check the form of this limit. As x approaches 0, tan(x) - x approaches 0 - 0 = 0, and x³ approaches 0, giving us a 0/0 indeterminate form. We can apply L'Hôpital's Rule by differentiating the numerator and denominator. The derivative of tan(x) - x is sec²(x) - 1, and the derivative of x³ is 3x². So the limit becomes lim(x→0) (sec²(x) - 1)/(3x²). This is still a 0/0 form, so we apply L'Hôpital's Rule again. The derivative of sec²(x) - 1 is 2sec²(x)tan(x), and the derivative of 3x² is 6x. So the limit becomes lim(x→0) (2sec²(x)tan(x))/(6x) = lim(x→0) (sec²(x)tan(x))/(3x). This is still a 0/0 form, so we apply L'Hôpital's Rule a third time. The derivative of sec²(x)tan(x) is sec²(x)(sec²(x) + 2tan²(x)), and the derivative of 3x is 3. So the limit becomes lim(x→0) sec²(x)(sec²(x) + 2tan²(x))/3. As x approaches 0, sec(x) approaches 1 and tan(x) approaches 0, so this becomes (1)(1 + 0)/3 = 1/3.

This limit is not in fraction form, so we need to rewrite it to apply L'Hôpital's Rule. We can rewrite x·ln(x) as ln(x)/(1/x). As x approaches 0 from the right, ln(x) approaches -∞ and 1/x approaches ∞, giving us a -∞/∞ indeterminate form. We can apply L'Hôpital's Rule by differentiating the numerator and denominator. The derivative of ln(x) is 1/x, and the derivative of 1/x is -1/x². So the limit becomes lim(x→0+) (1/x)/(-1/x²) = lim(x→0+) -x. As x approaches 0 from the right, -x approaches 0. Therefore, the limit equals 0.

This limit is not in fraction form, so we need to rewrite it to apply L'Hôpital's Rule. We can multiply the numerator and denominator by the conjugate: (√(x² + 1) - x) × (√(x² + 1) + x)/(√(x² + 1) + x) = (x² + 1 - x²)/(√(x² + 1) + x) = 1/(√(x² + 1) + x). As x approaches infinity, this becomes 1/∞ which implies a limit of 0.

First, we check the form of this limit. As x approaches 0, e^x - e^(-x) approaches e^0 - e^0 = 1 - 1 = 0, and sin(x) approaches 0, giving us a 0/0 indeterminate form. We can apply L'Hôpital's Rule by differentiating the numerator and denominator. The derivative of e^x - e^(-x) is e^x + e^(-x), and the derivative of sin(x) is cos(x). So the limit becomes lim(x→0) (e^x + e^(-x))/cos(x). As x approaches 0, e^x approaches 1, e^(-x) approaches 1, and cos(x) approaches 1. So the limit equals (1 + 1)/1 = 2.

First, we check the form of this limit. As x approaches 1, x³ - 1 approaches 1 - 1 = 0, and ln(x) approaches ln(1) = 0, giving us a 0/0 indeterminate form. We can apply L'Hôpital's Rule by differentiating the numerator and denominator. The derivative of x³ - 1 is 3x², and the derivative of ln(x) is 1/x. So the limit becomes lim(x→1) (3x²)/(1/x) = lim(x→1) 3x³. As x approaches 1, 3x³ approaches 3(1)³ = 3. Therefore, the limit equals 3.

This limit is not in fraction form and has the variable in the exponent. We can take the natural logarithm of the expression to simplify it. Let L = lim(x→0+) (1 + x)^(1/x). Then ln(L) = lim(x→0+) ln((1 + x)^(1/x)) = lim(x→0+) (1/x)·ln(1 + x) = lim(x→0+) ln(1 + x)/x. This is a 0/0 indeterminate form, so we can apply L'Hôpital's Rule. The derivative of ln(1 + x) is 1/(1 + x), and the derivative of x is 1. So ln(L) = lim(x→0+) (1/(1 + x))/1 = lim(x→0+) 1/(1 + x) = 1. Therefore, L = e^1 = e.

First, we check the form of this limit. As x approaches infinity, ln(x² + 1) approaches infinity and ln(x) approaches infinity, giving us an ∞/∞ indeterminate form. We can apply L'Hôpital's Rule by differentiating the numerator and denominator. The derivative of ln(x² + 1) is (2x)/(x² + 1), and the derivative of ln(x) is 1/x. So the limit becomes lim(x→∞) ((2x)/(x² + 1))/(1/x) = lim(x→∞) (2x²)/(x² + 1). This is still an ∞/∞ form, so we can apply L'Hôpital's Rule again. The derivative of 2x² is 4x, and the derivative of x² + 1 is 2x. So the limit becomes lim(x→∞) 4x/2x = lim(x→∞) 2 = 2.

First, we check the form of this limit. As x approaches 0, e^(2x) - 1 - 2x approaches e^0 - 1 - 0 = 1 - 1 - 0 = 0, and x² approaches 0, giving us a 0/0 indeterminate form. We can apply L'Hôpital's Rule by differentiating the numerator and denominator. The derivative of e^(2x) - 1 - 2x is 2e^(2x) - 2, and the derivative of x² is 2x. So the limit becomes lim(x→0) (2e^(2x) - 2)/(2x) = lim(x→0) (e^(2x) - 1)/x. This is still a 0/0 form, so we apply L'Hôpital's Rule again. The derivative of e^(2x) - 1 is 2e^(2x), and the derivative of x is 1. So the limit becomes lim(x→0) 2e^(2x)/1 = 2e^0 = 2(1) = 2.

L'Hôpital's Rule is a sufficient condition, not a necessary one. If the limit of f'(x)/g'(x) exists, the original limit equals it. However, if the limit of f'(x)/g'(x) does not exist (for example, if it oscillates), L'Hôpital's Rule is inconclusive. The original limit might still exist and would need to be evaluated using a different method, such as the Squeeze Theorem.

First, we check the form of this limit as x approaches infinity. The numerator x² + 3x approaches infinity, and the denominator 2x + 1 also approaches infinity, giving us an ∞/∞ indeterminate form. L'Hôpital's Rule can be applied directly to limits of this form without any algebraic manipulation. We can differentiate the numerator to get 2x + 3 and the denominator to get 2, then evaluate lim(x→∞) (2x + 3)/2, which equals infinity. Since the original limit is of the form ∞/∞, L'Hôpital's Rule can be directly applied, making the statement true.

First, we check the form of this limit. As x approaches 0, sin(x) - x + x³/6 approaches 0 - 0 + 0 = 0, and x⁵ approaches 0, giving us a 0/0 indeterminate form. We can apply L'Hôpital's Rule by differentiating the numerator and denominator. The derivative of sin(x) - x + x³/6 is cos(x) - 1 + x²/2, and the derivative of x⁵ is 5x^4. This is still a 0/0 form, so we continue applying L'Hôpital's Rule. After the second derivative, we get (-sin(x) + x)/(20x³). After the third derivative, we get (-cos(x) + 1)/(60x²). After the fourth derivative, we get sin(x)/(120x). After the fifth derivative, we get cos(x)/120. As x approaches 0, cos(x) approaches 1, so the limit equals 1/120.

To determine if L'Hôpital's Rule applies, we need to check if the limit produces an indeterminate form. As x approaches 0, the numerator x² + 1 approaches 1, and the denominator x approaches 0, giving us 1/0, which is not an indeterminate form. L'Hôpital's Rule specifically applies only to the indeterminate forms 0/0 or ∞/∞. Since this limit is not of either form, L'Hôpital's Rule cannot be applied. Instead, we recognize this as a limit that approaches infinity (or negative infinity depending on direction), not an indeterminate form. Therefore, the statement is false.