Exponential & Logarithmic Limits with L’Hôpital

Substitute x=0: tan(0)=0, denominator=0, so 0/0. Apply L'Hôpital's Rule. Let f(x)=tan(5x), f'(x)=5sec²(5x). g(x)=x, g'(x)=1. New limit: limit as x approaches 0 of 5sec²(5x)/1 = 5sec²(0)=5*1=5. Therefore, the original limit is 5.

Substitute x=0: numerator e^0 - e^0 = 1-1=0, denominator sin(0)=0, so 0/0. Apply L'Hôpital's Rule. f(x)=e^(3x)-e^x, f'(x)=3e^(3x)-e^x. g(x)=sin(2x), g'(x)=2cos(2x). New limit: limit as x approaches 0 of (3e^(3x)-e^x)/(2cos(2x)). Evaluate at x=0: (3*1 - 1)/(2*1) = (3-1)/2 = 2/2 = 1. Therefore, the original limit is 1.

As x approaches infinity, both numerator and denominator approach infinity. So we have infinity/infinity. Apply L'Hôpital's Rule. f(x)=ln(2x), f'(x)= (1/(2x))*2 = 1/x (using chain rule: derivative of ln(2x) is 1/(2x)*2 = 1/x). g(x)=ln(3x), g'(x)= (1/(3x))*3 = 1/x. So the new limit is limit as x approaches infinity of (1/x) / (1/x) = lim_x→infinity 1 = 1. Therefore, the original limit is 1.

Substitute x=0: numerator √(1)-1=0, denominator=0, so 0/0. Apply L'Hôpital's Rule. f(x)=√(1+x)-1 = (1+x)^(1/2)-1, f'(x)= (1/2)(1+x)^(-1/2). g(x)=x, g'(x)=1. New limit: limit as x approaches 0 of (1/2)(1+x)^(-1/2) / 1 = (1/2)(1+0)^(-1/2)= (1/2)*1 = 1/2. Therefore, the original limit is 1/2.

L'Hôpital's Rule states that if the limit of f(x)/g(x) as x approaches a is of the form 0/0 or infinity/infinity, and if the limit of f'(x)/g'(x) as x approaches a exists (or is infinite), then the original limit equals that limit. The derivatives do not need to be continuous at a, nor do they need to be defined at a (they need to exist near a, but not necessarily at a). They also do not need to be non-zero at a. The key condition is that the limit of the derivative ratio must exist or be infinite.

First, substitute x=0. The numerator is sin(0) - 0 = 0 and the denominator is 0³ = 0. This yields the indeterminate form 0/0, so we apply L'Hôpital's Rule. We differentiate the numerator and denominator repeatedly until a determinate form is reached. The derivative of the numerator is 5cos(5x) - 5, and the denominator is 3x². At x=0, this is still 0/0. Differentiating again, the numerator becomes -25sin(5x) and the denominator becomes 6x. At x=0, this is still 0/0. Differentiating one last time, the numerator becomes -125cos(5x) and the denominator becomes 6. Now, we can evaluate the limit directly: -125cos(0) / 6 = -125(1) / 6 = -125/6.

As x approaches infinity, numerator goes to infinity, denominator goes to infinity (since exponential growth). So we have infinity/infinity. Apply L'Hôpital's Rule. f(x)=x², f'(x)=2x. g(x)=2^x, g'(x)=2^x ln(2). New limit: limit as x approaches infinity of (2x)/(2^x ln(2)). This is still infinity/infinity? Actually, numerator goes to infinity, denominator goes to infinity. Apply L'Hôpital's Rule again. f2(x)=2x, f2'(x)=2. g2(x)=2^x ln(2), g2'(x)=2^x (ln(2))². New limit: limit as x approaches infinity of 2/(2^x (ln(2))²). As x approaches infinity, 2^x grows without bound, so the denominator goes to infinity, and the limit is 0. Therefore, the original limit is 0.

Direct substitution gives csc(0) - cot(0) which is undefined (infinity - infinity). So it's an indeterminate form of type infinity - infinity. We can combine the terms into a single fraction: csc(x) - cot(x) = 1/sin(x) - cos(x)/sin(x) = (1 - cos(x))/sin(x). Now as x approaches 0, numerator approaches 1-1=0 and denominator approaches 0, so it becomes 0/0. Then we can apply L'Hôpital's Rule. Therefore, option B is correct.

Substitute x=0: numerator sin(0)-0=0, denominator=0, so 0/0. Apply L'Hôpital's Rule. f(x)=sin(x)-x, f'(x)=cos(x)-1. g(x)=x³, g'(x)=3x². New limit: limit as x approaches 0 of (cos(x)-1)/(3x²). At x=0, this is (1-1)/0=0/0. Apply L'Hôpital's Rule again. f2(x)=cos(x)-1, f2'(x)=-sin(x). g2(x)=3x², g2'(x)=6x. New limit: limit as x approaches 0 of (-sin(x))/(6x) = limit as x approaches 0 of (-1/6) * (sin(x)/x). The limit of sin(x)/x as x approaches 0 is 1. So the limit becomes (-1/6)*1 = -1/6. Therefore, the original limit is -1/6.

The limit is of the form 0/0. L'Hôpital's Rule says to take the derivative of the numerator and the derivative of the denominator separately, then take the limit of that new fraction. So we differentiate numerator: derivative of e^x - 1 is e^x. Differentiate denominator: derivative of x is 1. Then the limit becomes limit as x approaches 0 of e^x/1 = e^0 = 1. Option B is incorrect because we do not use the quotient rule on the original fraction; we differentiate numerator and denominator separately. Option C is false because it is indeterminate. Option D is unnecessary because after one application we get a determinate form.

As x approaches 0 from the right, ln(x) approaches -∞, and cot(x) = cos(x)/sin(x) approaches infinity (since sin(x) is positive and small, cos(x) approaches 1). So we have -∞/infinity, an indeterminate form. Apply L'Hôpital's Rule. f(x)=ln(x), f'(x)=1/x. g(x)=cot(x), g'(x)=-csc²(x). New limit: limit as x approaches 0+ of (1/x) / (-csc²(x)) = limit as x approaches 0+ of -sin²(x)/x. We can evaluate this: as x approaches 0, sin²(x) ~ x², so sin²(x)/x ~ x, which goes to 0. So the limit is 0. Alternatively, apply L'Hôpital's Rule on -sin²(x)/x: limit as x approaches 0+ of -2sin(x)cos(x)/1 = 0. Therefore, the original limit is 0.

L'Hôpital's Rule states that if the limit of f'(x)/g'(x) is infinity (or negative infinity), then the original limit of f(x)/g(x) is also infinity (or negative infinity). So we can conclude the original limit is infinity.

As x approaches infinity, both numerator and denominator approach infinity. This is infinity/infinity. We can apply L'Hôpital's Rule repeatedly. Each application will reduce the power in the numerator by 1 while the denominator remains an exponential. After applying L'Hôpital's Rule 100 times, the numerator will become 100! (factorial) and the denominator will still be e^x. So the limit is limit as x approaches infinity of 100! / e^x = 0. Therefore, the original limit is 0.

First, we check if this limit produces an indeterminate form. As x approaches 0, sin(3x) approaches 0 and 2x approaches 0, so we have a 0/0 indeterminate form. This means we can apply L'Hôpital's Rule. We differentiate the numerator and denominator separately. The derivative of sin(3x) is 3cos(3x), and the derivative of 2x is 2. So the limit becomes lim(x→0) (3cos(3x))/2. Now we can directly substitute x = 0: (3cos(0))/2 = (3×1)/2 = 3/2.

First, we check the form of this limit. As x approaches infinity, ln(x) approaches infinity and 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(x) is 1/x, and the derivative of x is 1. So the limit becomes lim(x→∞) (1/x)/1 = lim(x→∞) 1/x. As x approaches infinity, 1/x approaches 0. Therefore, the limit equals 0.