MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: Indeterminate forms and L'Hôpital's Rule (MATH 203)

This sheet is not a summary of section 4.5 of Thomas' Calculus: you already have the course notes. It answers one question, what makes students lose marks on indeterminate forms and L'Hôpital's Rule in MATH 203 at Concordia University, and which precise gesture avoids each loss.

The rule is the easy part. The losses are in the algebra that surrounds it, the same algebra the weekly tutorials of the course drill: negative exponents, fractions of fractions, common denominators, laws of logarithms. Every trap below is one of those gestures done wrong inside a limit, with the sentence that earns the marks.

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The thread of the chapter

L'Hôpital's Rule is one line; the marks are lost in the algebra around it: a factor sent across the fraction bar flips the sign of its exponent, the complex fraction of each round is cleared before the next form is read, and a logarithm taken on the way down is undone with $e^L$ on the way back.

This chapter is part of MATH 203, Calculus I (Concordia)

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'Hôpital's Rule
Indeterminate form
Indeterminate quotients
Indeterminate products
Indeterminate differences
Indeterminate powers
Complex fractions
Negative exponents
Laws of logarithms
Relative rates of growth
Continuity of the exponential function

The essentials

The rule, as Thomas states it

• $f(a) = g(a) = 0$ (or both tend to $\pm\infty$ ), $f$ and $g$ differentiable near $a$ , and $g'(x) \ne 0$ near $a$ for $x \ne a$ .
• THEN $\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}$ , provided the limit on the right exists (or is $\pm\infty$ ).
• It holds for one-sided limits and for $x \to \pm\infty$ . The top and the bottom are differentiated SEPARATELY: no quotient rule.
• 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.
• A scientific calculator computes no limit. A value at a moderate $x$ is a check; a value at $x = 10^{-8}$ is roundoff.

Both curves vanish at

x = 1

, with slopes

\frac{1}{2}

and

1

: near

1

their quotient is close to

\frac{1}{2}

, and

\lim_{x\to 1}\frac{\sqrt x - 1}{\ln x} = \frac{1}{2}

Write the form, in symbols, on the line before each round. It is the permit, and it is also where the marker looks first.

The three algebra moves that carry the marks

• Across the bar, the exponent changes sign: $x^{a} = \frac{1}{x^{-a}}$ , so $x^{2/3}\ln x = \frac{\ln x}{x^{-2/3}}$ .
• After each round, clear the fraction of fractions: $\frac{1/x}{-\frac{2}{3}x^{-5/3}} = -\frac{3}{2}x^{-1 + 5/3} = -\frac{3}{2}x^{2/3}$ . Dividing by $x^{n}$ SUBTRACTS $n$ from the exponent.
• For a variable power: $\ln\left(f^{g}\right) = g\ln f$ , the exponent comes DOWN as a factor; at the end, $y = e^{\ln y}$ .
• For a difference: one common denominator, or factor out the dominant term: $x^{1/3} - \ln x = x^{1/3}\left(1 - \frac{\ln x}{x^{1/3}}\right)$ .
• Forms: $\frac{0}{0}$ and $\frac{\infty}{\infty}$ take the rule; $0 \cdot \infty$ and $\infty - \infty$ are rewritten first; $1^\infty$ , $0^0$ , $\infty^0$ go through the logarithm. $\frac{0}{c}$ , $\frac{c}{0}$ , $0^{\infty}$ are NOT indeterminate.

To test a clearing, plug one number into both sides: at $x = 8$ , $\frac{1/8}{-\frac{2}{3} \cdot 8^{-5/3}} = -6$ and $-\frac{3}{2} \cdot 8^{2/3} = -6$ .

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.

Rewrites around the rule: legal and illegal

Read a line as: the expression of the first column, rewritten as in the second, gives the result of the third. The red lines are rewrites that do not exist: they look like algebra and change the value.

Expression	Rewritten as	Result
$x^{2/3}\ln x$ at $0^+$	$\frac{\ln x}{x^{-2/3}}$	form $\frac{-\infty}{\infty}$ , limit $0$
Example: At $x = 0.001$ : $0.01 \times (-6.9078) \approx -0.0691$ , already close to $0$ .
$x^{2/3}\ln x$ at $0^+$	$\frac{\ln x}{x^{2/3}}$	$-\infty$ rule that does not exist
Example: At $x = 0.001$ : $\frac{-6.9078}{0.01} \approx -690.8$ , while the product is $-0.0691$ . What to do: A factor crosses the fraction bar with the sign of its exponent flipped: $x^{2/3} = \frac{1}{x^{-2/3}}$ .
$\frac{1/x}{-\frac{2}{3}x^{-5/3}}$	$-\frac{3}{2}x^{-1}x^{5/3}$	$-\frac{3}{2}x^{2/3}$
Example: At $x = 8$ : $\frac{1/8}{-1/48} = -6$ and $-\frac{3}{2} \cdot 4 = -6$ .
$\frac{1/x}{-\frac{2}{3}x^{-5/3}}$	$-\frac{2}{3}x^{-1}x^{-5/3}$	$-\frac{2}{3}x^{-8/3}$ rule that does not exist
Example: At $x = 8$ : $-\frac{2}{3} \cdot \frac{1}{256} \approx -0.0026$ , not $-6$ . What to do: Dividing by $x^{-5/3}$ is multiplying by $x^{5/3}$ : the exponents are subtracted, $-1 - \left(-\frac{5}{3}\right) = \frac{2}{3}$ .
$\ln\left(f^{g}\right)$	$g\ln f$	a product, form $0 \cdot \infty$ or $\infty \cdot 0$
Example: $\ln\left(2^{3}\right) = 3\ln 2 \approx 2.0794 = \ln 8$ .
$\ln\left(f^{g}\right)$	$(\ln f)^{g}$	a wrong value rule that does not exist
Example: $(\ln 2)^{3} \approx 0.333$ , while $\ln 8 \approx 2.079$ . What to do: The exponent comes DOWN as a factor: $\ln\left(f^{g}\right) = g \cdot \ln f$ .
$\frac{1}{x} - \frac{1}{xe^{x}}$	$\frac{e^{x} - 1}{xe^{x}}$	form $\frac{0}{0}$ at $0$ , limit $1$
Example: At $x = 0.01$ both sides give about $0.99502$ .

Each red line has the same SHAPE as the green line above it, which is exactly why it survives until a number is plugged in.

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. Using the rule on a form that is not indeterminate

the whole question

What not to write

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

What to write

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

Why: The rule needs $f(a) = g(a) = 0$ or two infinite limits. Without that hypothesis the derivatives compare slopes that have nothing to do with the values, and the page looks correct while being wrong.

2. Differentiating the two factors of a product

the whole question

What not to write

“ $\lim_{x\to 0^+} x^2\ln x = \lim_{x\to 0^+} 2x \cdot \frac{1}{x} = 2$ .”

What to write

“Form $0 \cdot (-\infty)$ . $x^2\ln x = \frac{\ln x}{x^{-2}}$ , form $\frac{-\infty}{\infty}$ ; one round gives $\frac{x^{-1}}{-2x^{-3}} = -\frac{x^{2}}{2} \to 0$ .”

Why: The rule is a statement about a QUOTIENT. A product must first become one, and the value $2$ that the faulty line produces is not even close: the product tends to $0$ .

3. Sending a factor downstairs without flipping its exponent

the whole limit, 3 to 4 marks

What not to write

“ $x^{2/3}\ln x = \frac{\ln x}{x^{2/3}} \to \frac{-\infty}{0^+} = -\infty$ .”

What to write

“ $x^{2/3}\ln x = \frac{\ln x}{x^{-2/3}}$ , form $\frac{-\infty}{\infty}$ ; one round gives $-\frac{3}{2}x^{2/3} \to 0$ .”

The product

x^{2/3}\ln x

(blue) returns to

0

at the origin, while the faulty rewrite

\frac{\ln x}{x^{2/3}}

(orange) dives to

-\infty

: two different functions.

Why: Moving $x^{2/3}$ under the bar DIVIDES by it; to keep the value, what goes under the bar is $x^{-2/3}$ . The faulty line computes the limit of a different function.

4. Clearing the fraction of fractions upside down

the whole limit

What not to write

“ $\lim_{x\to 0^+} \frac{\ln x}{\cot x} = \lim_{x\to 0^+} \frac{1/x}{-\csc^2 x} = \lim_{x\to 0^+} \left(-\frac{x}{\sin^2 x}\right) = -\infty$ .”

What to write

“ $\frac{1/x}{-\csc^2 x} = \frac{1}{x} \cdot \left(-\sin^2 x\right) = -\sin x \cdot \frac{\sin x}{x} \to 0 \cdot 1 = 0$ .”

Why: Dividing by $\csc^2 x$ is multiplying by $\sin^2 x$ , and $\frac{1}{x}$ stays where it is. Flipping the wrong level turns a limit of $0$ into an infinite one.

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 $\frac{1}{xe^{x}} \to \infty$ , so $\frac{1}{x} - \frac{1}{xe^{x}} \to 0$ .”

What to write

“Common denominator: $\frac{e^{x} - 1}{xe^{x}}$ , form $\frac{0}{0}$ ; one round gives $\frac{e^{x}}{e^{x} + xe^{x}} = \frac{1}{1 + x} \to 1$ .”

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

“ $3 - 2x \to 1$ and $1$ to any power is $1$ , so $\lim_{x\to 1} (3 - 2x)^{1/(x - 1)} = 1$ .”

What to write

“Form $1^{\pm\infty}$ . $\ln y = \frac{\ln(3 - 2x)}{x - 1}$ , form $\frac{0}{0}$ ; one round gives $\frac{-2}{3 - 2x} \to -2$ , so $y \to e^{-2}$ .”

Why: The base and the exponent move together; freezing the base at $1$ while the exponent grows is not a limit law. The logarithm follows both.

7. Giving the limit of the logarithm as the answer

1 to 2 marks

What not to write

“ $\ln y = \frac{\ln(2 - \cos 3x)}{x^2} \to \frac{9}{2}$ , so $\lim_{x\to 0} (2 - \cos 3x)^{1/x^2} = \frac{9}{2}$ .”

What to write

“ $\ln y \to \frac{9}{2}$ and the exponential is continuous, so $y = e^{\ln y} \to e^{9/2}$ .”

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

8. One more round after the form became c over 0

the whole question

What not to write

“ $\lim_{x\to 0} \frac{x + 1 - e^{x}}{x^3} = \lim \frac{1 - e^{x}}{3x^2} = \lim \frac{-e^{x}}{6x} = \lim \frac{-e^{x}}{6} = -\frac{1}{6}$ .”

What to write

“After two rounds the form is $\frac{-1}{0}$ : $-\infty$ as $x \to 0^+$ , $+\infty$ as $x \to 0^-$ . The limit does not exist.”

The two branches leave in opposite directions at

x = 0

: no finite limit, and certainly not

-\frac{1}{6}

Why: A form $\frac{c}{0}$ with $c \ne 0$ is concluded with the SIGNS of each side, never differentiated. The third round had no permit and invented a finite limit.

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: $\frac{0}{c}$ , $\frac{c}{0^{\pm}}$ , $\frac{c}{\infty}$ , $0^{\infty}$ → conclude directly, with the signs of each side; never differentiate

Example: $\frac{\cos x - x}{x + 2} \to \frac{1}{2}$ at $0$ ; $\frac{x^2 - 1}{2\ln x} \to 0$ at $0^+$
If $\frac{0}{0}$ or $\frac{\infty}{\infty}$ → one round, CLEAR the fraction of fractions, cancel, then write the new form

Example: $\frac{\ln(\cos x)}{\tan x} \to -\sin x\cos x \to 0$ at $\left(\frac{\pi}{2}\right)^-$

a common factor that cancels saves a whole round
If $0 \cdot \infty$ → send one factor downstairs with its exponent flipped; keep the logarithm upstairs

Example: $x(\ln x)^2 = \frac{(\ln x)^2}{x^{-1}} \to 0$ at $0^+$
If $\infty - \infty$ → common denominator, or factor out the dominant term

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

Example: $x^{3/(x - 1)} \to e^{3}$ at $x = 1$
If the rule loops, or each round is worse → algebra: divide by the dominant term, rewrite in sin and cos, or use a law of exponents

Example: $\frac{e^x + e^{-x}}{e^x - e^{-x}} \to 1$ ; $\frac{e^x}{e^{x^2}} = e^{x - x^2} \to 0$

The rule never applies to a product, a difference or a power directly. Three of the six branches start with algebra, not with a derivative.

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 a product of the form 0 times infinity

When to use it: Any limit of $f(x)\,g(x)$ with one factor tending to $0$ and the other to $\pm\infty$

1 Substitute and name the form: $0 \cdot \infty$ , with the signs.
2 Choose the factor to send downstairs: the power, never the logarithm. Write it with its exponent flipped: $x^{a} = \frac{1}{x^{-a}}$ .
3 Name the form of the new quotient, $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .
4 Apply the rule, naming the chain rule on any inner function, then clear the fraction of fractions, subtracting exponents.
5 Read the limit, or write the new form and repeat.

Concluding sentence

“Form $0 \cdot (-\infty)$ . Write $x^{2/3}\ln x = \frac{\ln x}{x^{-2/3}}$ , of the form $\frac{-\infty}{\infty}$ . By L'Hôpital's Rule, $\lim_{x\to 0^+}\frac{x^{-1}}{-\frac{2}{3}x^{-5/3}} = \lim_{x\to 0^+}\left(-\frac{3}{2}x^{2/3}\right) = 0$ .”

The trap: Writing $\frac{\ln x}{x^{2/3}}$ in the second line: the exponent did not change sign, and the limit found, $-\infty$ , is that of another function.

Marking: Typically 1 mark for the form, 2 for the rewritten quotient and its form, 3 for the round and the clearing, 1 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 is written above 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.

$\frac{-4\sin 2x}{2 + 6x}$ at $0$ is $\frac{0}{2}$ : a round after it is the error, and the $-\frac{4}{3}$ it produces is wrong; the limit is $0$ .
Every clearing survives one number

Plug a convenient value into the fraction of fractions and into your simplified line: they must agree. Five seconds, and it catches a fraction flipped the wrong way.

$\frac{1/x}{-\csc^2 x}$ at $x = 0.3$ is about $-0.2911$ , and so is $-\frac{\sin^2 x}{x}$ ; the faulty $-\frac{x}{\sin^2 x}$ gives about $-3.435$ .
A calculator check at a moderate value

Evaluate the original expression at a value close to the point but not absurdly close: roundoff destroys differences of nearly equal numbers.

$\frac{\tan x - \sin x}{x^3}$ gives $0.50002$ at $x = 0.01$ , which supports $\frac{1}{2}$ ; at $x = 10^{-5}$ a $10$ -digit calculator shows $0$ .
The sign and the size are possible

A power of a positive number is positive, so the answer to a 1 to the infinity is never the negative number you found for the logarithm.

$(\tan x)^{\tan 2x}$ near $\frac{\pi}{4}$ : $\ln y \to -1$ , and the answer is $e^{-1} \approx 0.37$ , not $-1$ .

The typical problem, taken apart

A power of the form 0 to the 0, all four algebra moves in one limit

Find $\lim_{x\to 1^+} (\ln x)^{x - 1}$ .

The calculator is allowed for a check, not for the answer. Every form must be named.

The base

\ln x

and the exponent

x - 1

both leave from

0

x = 1

: the form is

0^0

, and nothing can be read off yet.

Step 1

As $x \to 1^+$ , the base $\ln x \to 0^+$ and the exponent $x - 1 \to 0^+$ : form $0^0$ . Let $y = (\ln x)^{x - 1}$ , so $\ln y = (x - 1)\ln(\ln x)$ , of the form $0 \cdot (-\infty)$ .

Why

The exponent comes DOWN as a factor, $\ln\left(f^g\right) = g\ln f$ : the variable power becomes a product, which is the only thing the next step can handle.

Step 2

Send $x - 1$ downstairs with its exponent flipped: $\ln y = \frac{\ln(\ln x)}{(x - 1)^{-1}}$ , of the form $\frac{-\infty}{\infty}$ .

Why

The logarithm stays upstairs because its derivative is simpler than it; the power goes down as $(x - 1)^{-1}$ , not as $(x - 1)$ .

Step 3

Round 1: the top gives $\frac{1}{\ln x} \cdot \frac{1}{x}$ (chain rule, inner function $\ln x$ ), the bottom $-(x - 1)^{-2}$ . Clear: $\frac{1}{x\ln x} \cdot \left(-(x - 1)^{2}\right) = -\frac{(x - 1)^2}{x\ln x}$ , of the form $\frac{0}{0}$ .

Why

Dividing by $-(x - 1)^{-2}$ is multiplying by $-(x - 1)^{2}$ . Left uncleared, the next round would be a quotient rule on fractions.

Step 4

Round 2: the top gives $-2(x - 1)$ , the bottom $\ln x + 1$ (product rule). At $1$ : $\frac{0}{1}$ , not indeterminate, so $\ln y \to 0$ . Stop.

Why

The form is written again, and it is determinate: a third round would have no permit.

Step 5

Since the exponential is continuous, $y = e^{\ln y} \to e^{0} = 1$ . Check: $(\ln 1.0001)^{0.0001} \approx 0.9991$ .

Why

The answer is $e^{L}$ , not $L$ : answering $0$ here is the most frequent last-line error. The check uses a moderate value, and the approach is slow.

The conclusion, written out

“Form $0^0$ . With $y = (\ln x)^{x - 1}$ , $\ln y = \frac{\ln(\ln x)}{(x - 1)^{-1}}$ is of the form $\frac{-\infty}{\infty}$ ; two rounds of L'Hôpital's Rule, the second on a form $\frac{0}{0}$ , give $\ln y \to 0$ , so $\lim_{x\to 1^+} (\ln x)^{x - 1} = e^{0} = 1$ .”

The classic mistake on this problem: Answering $0$ because $\ln y \to 0$ , or answering $0$ because the base tends to $0$ : both confuse a form with a value.

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, concluded with the signs.
• Across the bar the exponent flips: $x^{a} = \frac{1}{x^{-a}}$ . Dividing by $x^{n}$ subtracts $n$ .
• $0 \cdot \infty$ : logarithm upstairs. $\infty - \infty$ : common denominator or dominant factor.
• $1^\infty$ , $0^0$ , $\infty^0$ : $\ln y = g\ln f$ , then $y \to e^{L}$ , never $L$ .
• $\ln x \ll x^{p} \ll e^{x}$ as $x \to \infty$ , for every $p > 0$ ; a calculator table does not prove a growth rate.

Frequently asked questions

When can I use L'Hôpital's Rule in MATH 203?

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 one or one over zero, conclude directly with the signs, without differentiating.

How do I simplify after applying L'Hôpital's Rule?

The new quotient is usually a fraction of fractions. Multiply the top by the reciprocal of the bottom, then combine the powers of x by subtracting exponents. Cancel any common factor before substituting. Test your simplified line by plugging one number into both versions: if they disagree, a fraction was flipped the wrong way.

How do I use L'Hôpital's Rule on zero times infinity?

First turn the product into a quotient by sending one factor under the fraction bar, and change the sign of its exponent when you do: x to the two thirds becomes one over x to the minus two thirds. Keep a logarithm on top, because its derivative is simpler than it. Then check the new form and apply the rule.

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

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.

Can I use my calculator to check a limit?

Yes, as a check, never as the answer. Evaluate the original expression at a value close to the point but not extremely close, such as 0.01 rather than 0.00001. Very close to the point, a scientific calculator subtracts nearly equal numbers and its roundoff can display zero for a limit that is one half.

Practise it

Corrected exercises: Indeterminate forms and L'Hôpital's Rule, MATH 203 at Concordia

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

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math203-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'Hôpital's Rule (MATH 203)

The essentials

The rule, as Thomas states it

The three algebra moves that carry the marks

The rules in table form

Rewrites around the rule: legal and illegal

The mistakes that cost marks

1. Using the rule on a form that is not indeterminate

2. Differentiating the two factors of a product

3. Sending a factor downstairs without flipping its exponent

4. Clearing the fraction of fractions upside down

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. One more round after the form became c over 0

Which method to choose

Which gesture, by the FORM of the limit

How the answer is expected to be written

Writing up a product of the form 0 times infinity

Check before you hand in

The typical problem, taken apart

A power of the form 0 to the 0, all four algebra moves in one limit

Learn by heart

Frequently asked questions

When can I use L'Hôpital's Rule in MATH 203?

How do I simplify after applying L'Hôpital's Rule?

How do I use L'Hôpital's Rule on zero times infinity?

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

Can I use my calculator to check a limit?

Corrected exercises: Indeterminate forms and L'Hôpital's Rule, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?