MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: indeterminate forms and L'Hospital's rule (MATH 140)

This sheet is not a summary of section 4.4 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on indeterminate forms and L'Hospital's rule in MATH 140 at McGill University, and which precise gesture avoids each loss.

The rule looks mechanical, and that is the danger: most errors are not in the derivatives but in using the rule where it has no permit. Every value below is exact and done by hand, as on the exam, and every trap comes with the sentence that earns the method mark.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next and tells you which chapter to tackle next. Create your space, an email is enough.

The thread of the chapter

L'Hospital's rule is a PERMIT checked at every use: the form must be $\frac{0}{0}$ or $\frac{\infty}{\infty}$ at THAT round, every other indeterminate form is first rewritten into one of these two, and the rule only concludes when $\lim \frac{f'}{g'}$ exists.

This chapter is part of MATH 140, Calculus 1 (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

L'Hospital's rule
Indeterminate form
Indeterminate quotients
Indeterminate products
Indeterminate differences
Indeterminate powers
Natural logarithm
Growth rates of functions
One-sided limits
Continuity of the exponential
Continuous compounding

The essentials

The rule, and its four hypotheses

• The quotient $\frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$ as $x \to a$ , with $a$ a number, $a^\pm$ , or $\pm\infty$ .
• $f$ and $g$ are differentiable near $a$ (not necessarily at $a$ ), and $g'(x) \ne 0$ near $a$ .
• $\lim \frac{f'(x)}{g'(x)}$ exists, or is $\pm\infty$ . THEN $\lim \frac{f(x)}{g(x)}$ is the same.
• Top and bottom are differentiated SEPARATELY. $\left(\frac{f}{g}\right)'$ has nothing to do with it.
• Why it works: near $a$ , $f(x) \approx f'(a)(x - a)$ and $g(x) \approx g'(a)(x - a)$ , so the quotient tends to the quotient of the SLOPES.

Both curves pass through the origin with slopes

3

and

1

: near

0

their quotient is close to

\frac{3x}{x}

, and

\lim_{x\to 0}\frac{\ln(1 + 3x)}{e^x - 1} = 3

Write the form, in symbols, on the line before each round: “form $\frac{0}{0}$ ”. It is the permit, and on a MATH 140 paper it is where the method mark is.

Seven indeterminate forms, and the forms that are not

• Quotients, the only ones the rule accepts: $\frac{0}{0}$ and $\frac{\infty}{\infty}$ .
• Product $0 \cdot \infty$ and difference $\infty - \infty$ : rewrite as a quotient first.
• Powers $1^\infty$ , $0^0$ , $\infty^0$ : take the logarithm, which gives a product $0 \cdot \infty$ .
• NOT indeterminate, concluded directly: $\frac{0}{c}$ with $c \ne 0$ gives $0$ ; $\frac{c}{0^\pm}$ with $c \ne 0$ gives $\pm\infty$ by the signs; $\infty + \infty = \infty$ ; $\infty \cdot \infty = \infty$ ; $0^\infty = 0$ .
• Growth as $x \to \infty$ : $\ln x \ll x^p \ll e^x$ for every $p > 0$ , where $u \ll v$ means $\frac{u}{v} \to 0$ .

A form is a statement about two functions, not a number: $\frac{0}{0}$ , $0^0$ or $1^\infty$ can end at any value, and the tables below show two of each ending differently.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Forms and their endings

Read a line as: an expression of the first column, in the situation of the second, has the limit of the third. The red lines are forms, not answers: the example and the counterexample have the same form and different limits.

Form	Situation	Limit
$\frac{0}{c}$ , $c \ne 0$	top $\to 0$ , bottom $\to c$	$0$
Example: $\frac{\ln x}{x + 1} \to \frac{0}{2} = 0$ as $x \to 1$ : no rule needed.
$\frac{c}{0^+}$ , $c > 0$	bottom $\to 0$ through positive values	$+\infty$
Example: $\frac{2 + \cos x}{x^2} \to \frac{3}{0^+} = +\infty$ as $x \to 0$ .
$0^\infty$	base $\to 0^+$ , exponent $\to \infty$	$0$
Example: $x^{1/x}$ as $x \to 0^+$ : $\ln y = \frac{\ln x}{x} \to \frac{-\infty}{0^+} = -\infty$ , so $y \to 0$ .
$\frac{0}{0}$	top and bottom $\to 0$	form $\frac{0}{0}$ settles nothing
Example: $\frac{5^x - 3^x}{x} \to \ln\frac{5}{3}$ as $x \to 0$ . Same form, other result: $\frac{x^2}{x} \to 0$ while $\frac{5x}{x} \to 5$ as $x \to 0$ . What to do: Check the form, then one round of the rule; or factor out $(x - a)$ .
$\frac{\infty}{\infty}$	top and bottom $\to \infty$	form $\frac{\infty}{\infty}$ settles nothing
Example: $\frac{\ln(x^3 + 1)}{\ln(x^2 + 5)} \to \frac{3}{2}$ as $x \to \infty$ . Same form, other result: $\frac{x}{e^x} \to 0$ while $\frac{e^x}{x} \to \infty$ . What to do: The rule, simplifying between rounds, or divide by the dominant term.
$0 \cdot \infty$	one factor $\to 0$ , the other $\to \pm\infty$	form $0 \cdot \infty$ settles nothing
Example: $\sin x \ln x = \frac{\ln x}{\csc x} \to 0$ as $x \to 0^+$ . Same form, other result: As $x \to 0^+$ : $x \cdot \frac{3}{x} = 3$ while $x \cdot \frac{1}{x^2} \to \infty$ . What to do: Write $\frac{f}{1/g}$ or $\frac{g}{1/f}$ , keeping the logarithm upstairs.
$\infty - \infty$	both terms $\to \infty$	form $\infty - \infty$ settles nothing
Example: $\frac{1}{\ln x} - \frac{1}{x - 1} \to \frac{1}{2}$ as $x \to 1^+$ . Same form, other result: $(x + 3) - x = 3$ while $x^2 - x \to \infty$ as $x \to \infty$ . What to do: Common denominator, factoring, or $\ln a - \ln b = \ln\frac{a}{b}$ .
$1^\infty$	base $\to 1$ , exponent $\to \infty$	form $1^\infty$ settles nothing
Example: $(\cos x)^{1/x^2} \to e^{-1/2}$ as $x \to 0$ . Same form, other result: $(1 + 2x)^{1/x} \to e^2$ while $(1 + x^2)^{1/x} \to 1$ as $x \to 0^+$ . What to do: $\ln y = g \ln f$ , a form $0 \cdot \infty$ ; then $y \to e^L$ .
$0^0$	base $\to 0^+$ , exponent $\to 0$	form $0^0$ settles nothing
Example: $x^{\sin x} \to 1$ as $x \to 0^+$ . Same form, other result: $x^{1/\ln x} = e$ for every $x$ in $(0, 1)$ , so its limit at $0^+$ is $e$ . What to do: Take the logarithm, as for $1^\infty$ .
$\infty^0$	base $\to \infty$ , exponent $\to 0$	form $\infty^0$ settles nothing
Example: $(e^x + x)^{1/x} \to e$ as $x \to \infty$ . Same form, other result: $(x^2)^{1/\ln x} = e^2$ for every $x > 1$ , while $(e^x + x)^{1/x} \to e$ . What to do: Take the logarithm, as for $1^\infty$ .

The first three lines are the ones students differentiate by reflex: they are not forms, and the rule applied to them gives wrong answers.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Differentiating a form that is not indeterminate

the whole question

What not to write

“ $\lim_{x\to 0} \frac{e^x + 1}{x + 1} = \lim_{x\to 0} \frac{e^x}{1} = 1$ by L'Hospital's rule.”

What to write

“At $0$ the form is $\frac{2}{1}$ , not indeterminate: $\lim_{x\to 0} \frac{e^x + 1}{x + 1} = \frac{2}{1} = 2$ .”

Why: The rule has a hypothesis, the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . Without it, the derivatives compare slopes that have nothing to do with the values, and the page looks perfectly correct while being wrong.

2. Using the quotient rule instead of two separate derivatives

the method marks of the question

What not to write

“ $\lim_{x\to 0} \frac{e^{3x} - 1}{x} = \lim_{x\to 0} \frac{3xe^{3x} - (e^{3x} - 1)}{x^2}$ .”

What to write

“Form $\frac{0}{0}$ . By L'Hospital's rule, $\lim_{x\to 0} \frac{3e^{3x}}{1} = 3$ .”

Why: The rule compares $f'$ with $g'$ , one at a time. The derivative of the quotient measures how the quotient CHANGES, not where it goes: here it tends to $\frac{9}{2}$ , not $3$ .

3. One round too many

the whole question

What not to write

“ $\lim_{x\to 0} \frac{1 - \cos x}{x^2 + 3x} = \lim_{x\to 0} \frac{\sin x}{2x + 3} = \lim_{x\to 0} \frac{\cos x}{2} = \frac{1}{2}$ .”

What to write

“Form $\frac{0}{0}$ ; one round gives $\frac{\sin x}{2x + 3}$ , of the form $\frac{0}{3}$ , not indeterminate. The limit is $0$ .”

Why: Each round needs its OWN permit. After the first round the denominator no longer tends to $0$ , and the second round compares slopes of functions that are not both vanishing.

4. Putting the wrong factor downstairs in a product

2 marks, and the time of the rest of the paper

What not to write

“ $x^2 e^x = \frac{e^x}{x^{-2}}$ , so by the rule $\frac{e^x}{-2x^{-3}} = -\frac{x^3 e^x}{2}$ , and I am stuck.”

What to write

“As $x \to -\infty$ , $x^2 e^x = \frac{x^2}{e^{-x}}$ , form $\frac{\infty}{\infty}$ ; two rounds give $\frac{2x}{-e^{-x}}$ , then $\frac{2}{e^{-x}} \to 0$ .”

Why: There are two ways to turn $fg$ into a quotient, and one of them makes the next quotient worse. Keep the power or the logarithm upstairs, where differentiating simplifies it, and the exponential downstairs.

5. Answering 0 to infinity minus infinity

2 to 3 marks, the whole limit

What not to write

“As $x \to 0^+$ , $\frac{1}{x} \to \infty$ and $\frac{1}{\ln(1 + x)} \to \infty$ , so $\frac{1}{x} - \frac{1}{\ln(1 + x)} \to 0$ .”

What to write

“Common denominator: $\frac{\ln(1 + x) - x}{x\ln(1 + x)}$ , form $\frac{0}{0}$ ; two rounds give $-\frac{1}{2}$ .”

Why: Two quantities that blow up at the same rate can differ by any amount; the common denominator turns the difference into a quotient that the rule can measure.

6. Answering 1 to a form 1 to the infinity

the whole limit

What not to write

“ $1 + 2x \to 1$ and $1$ to any power is $1$ , so $\lim_{x\to 0^+} (1 + 2x)^{1/x} = 1$ .”

What to write

“Form $1^\infty$ . $\ln y = \frac{\ln(1 + 2x)}{x}$ , form $\frac{0}{0}$ ; the rule gives $\frac{2}{1 + 2x} \to 2$ , so $y \to e^2$ .”

Three functions of the same form

1^\infty

0^+

reach three different heights:

e^2

1

and

e^{-1}

. The form alone decides nothing.

Why: The base and the exponent move at the same time. Freezing the base at $1$ while the exponent keeps growing is not a limit law; the logarithm keeps track of both.

7. Giving the limit of the logarithm as the answer

1 to 2 marks

What not to write

“ $\ln y = \cot x \ln(1 + \sin 3x) \to 3$ , so $\lim_{x\to 0^+} (1 + \sin 3x)^{\cot x} = 3$ .”

What to write

“ $\ln y \to 3$ and the exponential is continuous, so $y = e^{\ln y} \to e^3$ .”

Why: The logarithm is a detour, not the destination. The last line must come back through $e^{(\cdot)}$ , naming the continuity of the exponential.

8. Declaring that a limit does not exist because the rule was silent

the whole question

What not to write

“ $\frac{x - \cos x}{x + \cos x}$ : by L'Hospital's rule, $\frac{1 + \sin x}{1 - \sin x}$ has no limit, so the limit does not exist.”

What to write

“Divide by $x$ : $\frac{1 - \frac{\cos x}{x}}{1 + \frac{\cos x}{x}} \to \frac{1 - 0}{1 + 0} = 1$ , since $\left|\frac{\cos x}{x}\right| \le \frac{1}{x} \to 0$ .”

The quotient settles on the line

y = 1

while

g'(x) = 1 - \sin x

keeps touching

0

: the rule has no permit here, and the limit exists anyway.

Why: The rule is an implication: IF $\lim \frac{f'}{g'}$ exists THEN it equals $\lim \frac{f}{g}$ . When it does not, the rule says nothing, and here even its hypothesis $g' \ne 0$ fails, since $1 - \sin x$ vanishes infinitely often.

Which method to choose

Which gesture, by the FORM of the limit

Substitute the point first, write the form you get, and let the form pick the gesture

If not indeterminate: $\frac{0}{c}$ , $\frac{c}{0^\pm}$ , $\infty + \infty$ , $0^\infty$ → conclude directly, with the signs; never differentiate

Example: $\frac{\ln x}{x + 1} \to 0$ at $x = 1$ ; $\frac{e^x - 1}{x} \to 0$ at $-\infty$
If $\frac{0}{0}$ or $\frac{\infty}{\infty}$ → one round of the rule, then simplify, then write the new form

Example: $\frac{\ln x}{x^2 - 1} \to \frac{1/x}{2x} \to \frac{1}{2}$ at $x = 1$

for polynomials, factoring is just as fast and allowed when the rule is not
If $0 \cdot \infty$ → rewrite as a quotient, logarithm or power upstairs, exponential or trigonometric factor downstairs

Example: $x e^x = \frac{x}{e^{-x}} \to 0$ as $x \to -\infty$
If $\infty - \infty$ → common denominator, factor the dominant term, or combine logarithms

Example: $\sec x - \tan x = \frac{1 - \sin x}{\cos x} \to 0$ as $x \to \frac{\pi}{2}^-$
If a variable exponent: $1^\infty$ , $0^0$ , $\infty^0$ → take the logarithm, find its limit $L$ , answer $e^L$

Example: $(1 + \sin 4x)^{\cot x} \to e^4$ as $x \to 0^+$
If the rule loops or makes the quotient worse → divide by the dominant term, or substitute $t = \frac{1}{x}$

Example: $\frac{\sqrt{x^2 + 1}}{x} \to 1$ at $\infty$ ; $\frac{e^{-1/x}}{x} = \frac{t}{e^t} \to 0$

If $\lim \frac{f'}{g'}$ does not exist, go back to $\frac{f}{g}$ itself: squeeze, or divide by the dominant term. The silence of the rule is not an answer.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing up an indeterminate power

When to use it: Any limit of $f(x)^{g(x)}$ in which the base and the exponent both depend on $x$

1 Substitute and name the form: $1^\infty$ , $0^0$ or $\infty^0$ . If it is $0^\infty$ or $c^{\pm\infty}$ , conclude directly.
2 Set $y = f(x)^{g(x)}$ , check $f(x) > 0$ near the point, and write $\ln y = g(x)\ln f(x)$ .
3 Rewrite $\ln y$ as a quotient and name its form, $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .
4 Apply the rule, naming the chain rule on each inner function, and simplify until the limit $L$ of $\ln y$ can be read.
5 Conclude: the exponential is continuous, so $y = e^{\ln y} \to e^L$ .

Concluding sentence

“Form $1^\infty$ . Let $y = (\cos x)^{1/x^2}$ ; then $\ln y = \frac{\ln \cos x}{x^2}$ , of the form $\frac{0}{0}$ , and by L'Hospital's rule $\ln y \to \lim_{x\to 0} \frac{-\tan x}{2x} = -\frac{1}{2}$ . Since the exponential is continuous, $\lim_{x\to 0} (\cos x)^{1/x^2} = e^{-1/2}$ .”

The trap: Stopping at $\ln y \to -\frac{1}{2}$ and answering $-\frac{1}{2}$ : a power of a positive number is never negative.

Marking: Typically 1 mark for the form, 2 for the logarithm and the quotient, 4 for the rule and the limit of the logarithm, 1 for the continuity of the exponential, 2 for the final value.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The form was written before every round

Reread the page and look for the form, in symbols, above each quotient of derivatives. A round without its form is a round that may not have been allowed.

$\frac{\sin x}{2x + 3}$ at $0$ is $\frac{0}{3}$ : a round after it is the error, and the answer $\frac{1}{2}$ it produced is wrong.
The sign and the size are possible

Compare the answer with what the function can do: a power of a positive number is positive, a quotient of positive terms is not negative, a bounded function has no infinite limit.

$(\cos x)^{1/x^2}$ lies between $0$ and $1$ near $0$ , so $-\frac{1}{2}$ is impossible and $e^{-1/2} \approx 0.61$ is plausible.
The growth ranking agrees

At infinity, keep only the dominant term of each part, and compare with the ranking logarithm, power, exponential.

$\frac{\ln(x^3 + 1)}{\ln(x^2 + 5)}$ behaves like $\frac{3\ln x}{2\ln x}$ , so $\frac{3}{2}$ ; an answer of $\infty$ would say a logarithm beats a logarithm.
A second method on the easy cases

When factoring or a known tangent line also works, do it in the margin: the two answers must agree.

$\frac{x^3 - 8}{x^2 + x - 6}$ at $2$ : the rule gives $\frac{12}{5}$ , and $\frac{x^2 + 2x + 4}{x + 3}$ at $2$ gives $\frac{12}{5}$ too.

The typical problem, taken apart

A difference of two infinities that ends at one half

Find $\lim_{x\to 0} \left(\frac{1}{x} - \frac{1}{e^x - 1}\right)$ .

No calculator. Every form must be named, as on a MATH 140 final.

Both curves have a vertical asymptote at

x = 0

and blow up with the same sign on each side, yet they stay close to each other: their difference has a finite limit.

Step 1

As $x \to 0^+$ , $\frac{1}{x} \to +\infty$ and $\frac{1}{e^x - 1} \to +\infty$ ; as $x \to 0^-$ , both tend to $-\infty$ . On each side the form is $\infty - \infty$ .

Why

Checking BOTH sides shows that the two terms blow up together: had they had opposite signs, the limit would have been infinite and nothing more would be needed.

Step 2

Common denominator: $\frac{1}{x} - \frac{1}{e^x - 1} = \frac{e^x - 1 - x}{x(e^x - 1)}$ . At $0$ the top is $1 - 1 - 0 = 0$ and the bottom is $0$ : form $\frac{0}{0}$ .

Why

The rule never applies to a difference. The common denominator is the rewriting that gives it a quotient, and naming the new form is the permit.

Step 3

Round 1: $\frac{e^x - 1}{(e^x - 1) + xe^x}$ , by the product rule downstairs. At $0$ : $\frac{0}{0 + 0}$ , still $\frac{0}{0}$ .

Why

The product rule on $x(e^x - 1)$ is where the marks leak: $x \cdot e^x$ plus $1 \cdot (e^x - 1)$ . The form is checked again before going on.

Step 4

Round 2: $\frac{e^x}{e^x + e^x + xe^x} = \frac{e^x}{e^x(2 + x)} = \frac{1}{2 + x} \to \frac{1}{2}$ .

Why

Simplifying $e^x$ before substituting keeps the last line readable. The form $\frac{1}{2}$ is not indeterminate: stop here.

Step 5

Check the sign: for $x > 0$ , $e^x - 1 > x$ (the tangent line $y = x$ lies below $e^x - 1$ ), so $\frac{1}{e^x - 1} < \frac{1}{x}$ and the difference is positive, like $\frac{1}{2}$ .

Why

A five second check that catches a lost sign in the product rule, which would have produced $-\frac{1}{2}$ .

The conclusion, written out

“Form $\infty - \infty$ . With a common denominator, $\frac{e^x - 1 - x}{x(e^x - 1)}$ is of the form $\frac{0}{0}$ ; two rounds of L'Hospital's rule, each on a form $\frac{0}{0}$ , give $\frac{1}{2 + x}$ , so $\lim_{x\to 0}\left(\frac{1}{x} - \frac{1}{e^x - 1}\right) = \frac{1}{2}$ .”

The classic mistake on this problem: Answering $0$ because the two terms both tend to infinity, or applying the rule to $\frac{1}{x}$ and $\frac{1}{e^x - 1}$ separately, which has no meaning.

Learn by heart

• The rule: form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , $g' \ne 0$ near the point, $\lim \frac{f'}{g'}$ exists; then $\lim \frac{f}{g} = \lim \frac{f'}{g'}$ .
• Top and bottom separately, never the quotient rule. The form is written before EVERY round.
• $\frac{0}{c}$ , $\frac{c}{0}$ with $c \ne 0$ , $0^\infty$ : not indeterminate, never differentiated.
• $0 \cdot \infty$ : a quotient, logarithm upstairs. $\infty - \infty$ : common denominator or logarithm laws.
• $1^\infty$ , $0^0$ , $\infty^0$ : $\ln y = g\ln f$ , then $y \to e^L$ .
• $\ln x \ll x^p \ll e^x$ as $x \to \infty$ , for every $p > 0$ .
• If $\frac{f'}{g'}$ has no limit or the rule loops: algebra, never the conclusion that the limit does not exist.

Frequently asked questions

When can I use L'Hospital's rule?

Only on a quotient whose form, at that moment, is zero over zero or infinity over infinity, with both functions differentiable near the point and the derivative of the denominator not zero there. Write the form before each use. If the form is anything else, such as zero over two, conclude directly without differentiating.

Do I use the quotient rule with L'Hospital's rule?

No. Differentiate the numerator on its own and the denominator on its own, then take the limit of the new quotient. The quotient rule computes the derivative of the whole fraction, which describes how fast the fraction changes, not where it goes, and it gives a different number in general.

How do I use L'Hospital's rule on zero times infinity or infinity minus infinity?

First turn the expression into a quotient. For a product, divide one factor by the reciprocal of the other, keeping a logarithm or a power on top. For a difference, use a common denominator, factor out the dominant term, or combine two logarithms into one. Then check the new form and apply the rule.

How do I find a limit like one to the power infinity?

Call the expression y and take the natural logarithm: the exponent comes down and multiplies the logarithm of the base. Rewrite that product as a quotient, find its limit L with the rule, then answer e to the power L, because the exponential is continuous. Never answer L itself, and never answer 1 just because the base tends to 1.

What if L'Hospital's rule keeps going in circles?

Stop differentiating. When the rule gives back the quotient you started from, or a worse one, use algebra: divide top and bottom by the dominant term, remembering that the square root of x squared is the absolute value of x, or substitute t equal to one over x. The rule is a tool, not an obligation.

Practise it

Corrected exercises: L'Hospital's rule, MATH 140 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet The shape of a graph Next sheet Curve sketching

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math140-lhopital. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: indeterminate forms and L'Hospital's rule (MATH 140)

The essentials

The rule, and its four hypotheses

Seven indeterminate forms, and the forms that are not

The rules in table form

Forms and their endings

The mistakes that cost marks

1. Differentiating a form that is not indeterminate

2. Using the quotient rule instead of two separate derivatives

3. One round too many

4. Putting the wrong factor downstairs in a product

5. Answering 0 to infinity minus infinity

6. Answering 1 to a form 1 to the infinity

7. Giving the limit of the logarithm as the answer

8. Declaring that a limit does not exist because the rule was silent

Which method to choose

Which gesture, by the FORM of the limit

How the answer is expected to be written

Writing up an indeterminate power

Check before you hand in

The typical problem, taken apart

A difference of two infinities that ends at one half

Learn by heart

Frequently asked questions

When can I use L'Hospital's rule?

Do I use the quotient rule with L'Hospital's rule?

How do I use L'Hospital's rule on zero times infinity or infinity minus infinity?

How do I find a limit like one to the power infinity?

What if L'Hospital's rule keeps going in circles?

Corrected exercises: L'Hospital's rule, MATH 140 at McGill

See also

Looking for a MATH 140 tutor in Montreal?