Difficulty: Intermediate | Prerequisites: Core concepts study notes on L'Hopital's Rule and indeterminate forms
These are worked solutions for free-response limit problems that require L'Hopital's Rule or related techniques. In an exam setting, showing your work means identifying the indeterminate form, stating when and why you are applying L'Hopital's Rule, and simplifying to an exact answer (no decimal approximations). The problems here cover 0/0 and infinity/infinity quotients, 0 times infinity products, infinity minus infinity differences, and exponential indeterminate forms. Each solution follows the pattern: identify form, convert if necessary, differentiate, evaluate.
Direct substitution gives 0/0.
Apply L'Hopital's Rule: differentiate numerator and denominator separately.
d/d(theta)[arctan(theta)] = 1 / (1 + theta^2)
d/d(theta)[2 theta] = 2
The limit becomes lim (theta to 0) of [1 / (1 + theta^2)] / 2 = (1/1) / 2 = 1/2.
Answer: 1/2
This is not an indeterminate form. As theta approaches pi from the right, (theta minus pi) approaches 0 from the positive side, so sin(theta minus pi) approaches 0 through positive values.
Therefore 1 / sin(theta minus pi) goes to positive infinity.
L'Hopital's Rule does not apply here because the limit is of the form 1/0, which is not indeterminate.
Answer: positive infinity
Direct substitution gives 0/0.
Apply L'Hopital's Rule:
d/dx[2x minus 2 pi] = 2
d/dx[sin(x minus pi)] = cos(x minus pi)
The limit becomes 2 / cos(pi minus pi) = 2 / cos(0) = 2 / 1 = 2.
Answer: 2
As z approaches pi from the right, (z minus pi) approaches 0+, so 1/(z minus pi) approaches positive infinity.
sin of a quantity going to infinity oscillates between -1 and 1 and does not settle on a single value.
Answer: the limit does not exist
This is the form 0 times infinity. Rewrite cot(2 alpha) as cos(2 alpha) / sin(2 alpha).
The expression becomes alpha times cos(2 alpha) / sin(2 alpha).
As alpha goes to 0, cos(2 alpha) goes to 1, so we focus on alpha / sin(2 alpha), which is 0/0.
Apply L'Hopital's Rule to alpha / sin(2 alpha):
d/d(alpha)[alpha] = 1
d/d(alpha)[sin(2 alpha)] = 2 cos(2 alpha)
The limit becomes 1 / (2 cos(0)) = 1/2.
Multiply by the cos(2 alpha) factor (which went to 1): the answer is 1/2.
Answer: 1/2
This is the form infinity times 0 (since ln((y+1)/(y-1)) goes to ln 1 = 0 as y goes to infinity).
Rewrite as ln((y+1)/(y-1)) / (1/y), which is 0/0.
Simplify the logarithm: ln(y+1) minus ln(y-1).
Apply L'Hopital's Rule:
Numerator derivative: 1/(y+1) minus 1/(y-1) = [(y-1) minus (y+1)] / [(y+1)(y-1)] = -2 / (y^2 minus 1)
Denominator derivative: -1/y^2
The ratio is [-2 / (y^2 - 1)] / [-1/y^2] = 2y^2 / (y^2 - 1).
Divide top and bottom by y^2: 2 / (1 - 1/y^2). As y goes to infinity, this gives 2/1 = 2.
Answer: 2
As x approaches 0 from the left, 1/x approaches negative infinity, so e^(1/x) approaches 0. Meanwhile x^3 approaches 0.
This is the form 0 times 0, which is not indeterminate. The product of two quantities both going to zero is simply 0.
Answer: 0
This is the form 1^infinity. Let y = (1 + 3/t)^t.
Take ln: ln y = t times ln(1 + 3/t).
Rewrite as ln(1 + 3/t) / (1/t), which is 0/0 as t goes to infinity.
Apply L'Hopital's Rule:
Numerator derivative: [1/(1 + 3/t)] times (-3/t^2) = -3 / [t^2(1 + 3/t)]
Denominator derivative: -1/t^2
The ratio simplifies to 3 / (1 + 3/t). As t goes to infinity, this gives 3/1 = 3.
So lim ln y = 3, which means the original limit is e^3.
Answer: e^3
This is the form infinity minus infinity.
Factor out sqrt(w): sqrt(w) times [ln(w) / sqrt(w) minus 1].
Now evaluate lim (w to infinity) of ln(w) / sqrt(w). This is infinity/infinity.
Apply L'Hopital's Rule:
d/dw[ln w] = 1/w
d/dw[sqrt(w)] = 1/(2 sqrt(w))
The ratio becomes (1/w) / (1/(2 sqrt(w))) = 2 sqrt(w) / w = 2 / sqrt(w), which goes to 0.
So the bracket becomes [0 minus 1] = -1, and the whole expression is sqrt(w) times (-1), which goes to negative infinity.
Answer: negative infinity
Always check the form before reaching for L'Hopital's Rule. Problem 11 (1/0) and Problem 16 (0 times 0) do not require it at all.
Products (0 times infinity) need to be rewritten as quotients first. The choice of which factor goes in the denominator matters: pick whichever makes the derivative cleaner.
Exponential forms (1^infinity) always follow the same script: take ln, convert to a quotient, apply L'Hopital's Rule, then exponentiate.
For infinity minus infinity, factoring out a dominant term is often the fastest route.
On an exam, state the indeterminate form explicitly before applying L'Hopital's Rule. This is how you earn the justification marks.