Difficulty: Advanced | Prerequisites: Core concepts study notes, Part 1 solutions, differential equations, Euler's method
As t goes to infinity, sin(1/t) approaches sin(0) = 0, and ln t goes to infinity. The form is 0/infinity, which equals 0. This is not indeterminate.
Answer: 0
Rewrite e^(-v) = 1/e^v. So the expression is v^2 times e^v, which is infinity times infinity = infinity. This is not indeterminate.
Alternatively, written as v^2 / e^(-v) = v^2 times e^v. Both factors grow without bound.
Answer: positive infinity
This requires careful handling. sin(1/x) oscillates between -1 and 1 as x approaches 0, so x^2 times sin(1/x) is bounded between -x^2 and x^2, both of which go to 0.
Meanwhile sin x goes to 0 as x goes to 0. So we have the form 0/0, but L'Hopital's Rule is problematic here because the derivative of sin(1/x) does not behave well.
Instead, rewrite: [x^2 sin(1/x)] / sin x = [x^2 sin(1/x) / x] times [x / sin x].
The second factor, x / sin x, goes to 1.
The first factor is x times sin(1/x). Since sin(1/x) is bounded between -1 and 1, this is squeezed between -x and x, which both go to 0. By the Squeeze Theorem, x sin(1/x) goes to 0.
The product is 0 times 1 = 0.
Answer: 0
As u goes to 0+, 2^u minus 1 goes to 0 and sqrt(u) goes to 0. The form is 0 times 0, which is simply 0. No L'Hopital's Rule needed.
Answer: 0
Combine the logarithms: ln|(2x minus 4) / (x + 3)|.
Evaluate the inside: (2x - 4)/(x + 3). Divide numerator and denominator by x: (2 - 4/x) / (1 + 3/x). As x goes to infinity, this approaches 2/1 = 2.
So the limit is ln|2| = ln 2.
Answer: ln 2
Direct substitution gives 0/0 (the pi^2 is a constant factor, tan(2 theta) goes to 0, and theta goes to 0, while cos(2 theta) goes to 1).
Simplify: pi^2 times [tan(2 theta) / theta] times [1 / cos(2 theta)].
The factor 1/cos(2 theta) goes to 1/1 = 1.
For tan(2 theta) / theta: this is sin(2 theta) / (theta times cos(2 theta)). Apply L'Hopital's Rule to sin(2 theta) / theta (form 0/0):
d/d(theta)[sin(2 theta)] = 2 cos(2 theta)
d/d(theta)[theta] = 1
The limit is 2 cos(0) = 2.
Combining everything: pi^2 times 2 times 1 = 2 pi^2.
Answer: 2 pi^2
Given: dy/dx = 1 minus y, with f(1) = 0 and f(x) < 1 for all x.
Starting at x = 1, two steps of equal size to reach x = 0. Step size h = (0 minus 1)/2 = -1/2.
Step 1: At (1, 0), dy/dx = 1 minus 0 = 1. New y = 0 + (-1/2)(1) = -1/2. New point: (1/2, -1/2).
Step 2: At (1/2, -1/2), dy/dx = 1 minus (-1/2) = 3/2. New y = -1/2 + (-1/2)(3/2) = -1/2 minus 3/4 = -5/4. New point: (0, -5/4).
f(0) is approximately -5/4.
Since f(1) = 0 and x^3 minus 1 = 0 when x = 1, this is 0/0. Apply L'Hopital's Rule.
Numerator derivative: f'(x). From the differential equation, f'(x) = 1 minus f(x). At x = 1: f'(1) = 1 minus 0 = 1.
Denominator derivative: 3x^2. At x = 1: 3(1)^2 = 3.
The limit is 1/3.
dy/dx = 1 minus y is a separable equation.
Separate: dy / (1 minus y) = dx.
Integrate both sides: -ln|1 minus y| = x + C.
So ln|1 minus y| = -x minus C.
|1 minus y| = e^(-x minus C) = Ae^(-x) where A = e^(-C).
Since f(x) < 1 for all x, we have 1 minus y > 0, so 1 minus y = Ae^(-x).
Apply initial condition f(1) = 0: 1 minus 0 = Ae^(-1), so A = e.
1 minus y = e times e^(-x) = e^(1-x).
y = 1 minus e^(1-x).
f(x) = 1 minus e^(1 minus x)
Given: dy/dx = x^2 minus (1/2)y.
Differentiate dy/dx = x^2 minus (1/2)y with respect to x:
d^2y/dx^2 = 2x minus (1/2)(dy/dx) = 2x minus (1/2)(x^2 minus (1/2)y) = 2x minus (1/2)x^2 + (1/4)y.
d^2y/dx^2 = 2x minus (1/2)x^2 + (1/4)y
Check dy/dx at (-2, 8): (-2)^2 minus (1/2)(8) = 4 minus 4 = 0. So there is a critical point.
Check d^2y/dx^2 at (-2, 8): 2(-2) minus (1/2)(4) + (1/4)(8) = -4 minus 2 + 2 = -4.
Since d^2y/dx^2 < 0, the graph has a relative maximum at (-2, 8) by the Second Derivative Test.
At x = -1: g(-1) = 2, so the numerator is 0. The denominator 3(x+1)^2 is also 0. Form is 0/0.
Apply L'Hopital's Rule: g'(x) / [6(x + 1)].
At x = -1: g'(-1) = (-1)^2 minus (1/2)(2) = 1 minus 1 = 0. Still 0/0.
Apply L'Hopital's Rule again: g''(x) / 6.
At x = -1: g''(-1) = 2(-1) minus (1/2)(1) + (1/4)(2) = -2 minus 1/2 + 1/2 = -2.
The limit is -2/6 = -1/3.
Answer: -1/3
BC version (Euler's method): Starting at x = 0, two steps of size 1/2 to reach x = 1.
Step 1: At (0, 2), dy/dx = 0 minus (1/2)(2) = -1. New y = 2 + (1/2)(-1) = 3/2. New point: (1/2, 3/2).
Step 2: At (1/2, 3/2), dy/dx = 1/4 minus (1/2)(3/2) = 1/4 minus 3/4 = -1/2. New y = 3/2 + (1/2)(-1/2) = 3/2 minus 1/4 = 5/4. New point: (1, 5/4).
h(1) is approximately 5/4.
AB version (Linearisation): At x = 0, h(0) = 2 and h'(0) = 0 minus (1/2)(2) = -1.
The linearisation is L(x) = 2 + (-1)(x minus 0) = 2 minus x.
L(1) = 2 minus 1 = 1.
h(1) is approximately 1.
Not every limit needs L'Hopital's Rule. Problems 19, 20, and 22 resolve without it because the form is not indeterminate (0/infinity, infinity times infinity, 0 times 0). Identifying the form first saves time and avoids errors.
The Squeeze Theorem (Problem 21) is a valuable alternative when derivatives are messy or undefined. If an oscillating factor is bounded and multiplied by something going to zero, squeeze it.
Logarithm properties (Problem 23) let you sidestep L'Hopital's Rule entirely. When two logs are subtracted, combine them into one and evaluate the argument's limit.
AP free-response problems (25 and 26) test L'Hopital's Rule alongside differential equations. The key move is using dy/dx = F(x, y) to find f'(a) from the DE rather than solving the DE first. For part (c) of Problem 26, you need the second derivative, which means differentiating the DE with respect to x (using implicit differentiation via the chain rule on the y term).
Euler's method questions pair naturally with L'Hopital's Rule on the AP exam. Both appear in the same free-response problem, testing whether you can switch between numerical approximation and analytical limit evaluation.