Integral Calculus 201-SN3-RE (201-NYB) • CEGEP, Montreal

Revision sheet: indeterminate forms and L'Hôpital's rule, 201-SN3-RE (201-NYB)

This sheet does not prove L'Hôpital's rule again: it fits in one line. It deals with what decides the mark, namely recognizing the form BEFORE differentiating, and rewriting the five indeterminate forms that are not quotients, the only real difficulty of the chapter, as it comes back in Integral Calculus 201-SN3-RE (formerly 201-NYB).

It is written for CEGEP students in the Science program (Sciences de la nature). The corrected exercise set on the same chapter then puts each reflex to the test.

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The thread of the chapter

L'Hôpital's rule only applies to the forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$ . Every other indeterminate form must first be REWRITTEN as one of these two, and it is this rewriting, never the differentiation, that makes the question hard.

This chapter is part of 201-SN3-RE (201-NYB), Integral Calculus

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 forms
Quotient of the derivatives
Rewriting as a quotient
Indeterminate exponential form
Taking the logarithm
Comparison of growth rates
Limit of a sequence
Special trigonometric limits

The essentials

The exact statement, and its four conditions

• If $\lim\frac{f}{g}$ has the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , if $f$ and $g$ are differentiable near the point, if $g'$ does not vanish there, and if $\lim\frac{f'}{g'}$ exists (finite or infinite), then $\lim\frac{f}{g}=\lim\frac{f'}{g'}$ .
• Differentiate the NUMERATOR and the DENOMINATOR separately, never the quotient.
• The rule can be repeated as long as the form stays indeterminate; the form must be CHECKED AGAIN at every round.
• It applies to NO other form: $\frac{5}{0^{+}}$ , $\frac{0}{5}$ , $\frac{5}{\infty}$ are read off directly.

\frac{\sin x}{x}

does not exist at

0

, but the curve heads toward

1

from both sides: that is what the form

\frac{0}{0}

was hiding, and what L'Hôpital's rule finds in one line.

Writing “form $\frac{0}{0}$ ” before each application earns half a mark and prevents the most frequent mistake of the chapter.

The seven forms, and how to rewrite them

• $\frac{0}{0}$ and $\frac{\infty}{\infty}$ : L'Hôpital's rule directly.
• $0\cdot\infty$ : write $fg=\frac{f}{1/g}$ , which gives $\frac{0}{0}$ , or $\frac{g}{1/f}$ , which gives $\frac{\infty}{\infty}$ .
• $\infty-\infty$ : use a common denominator, factor out the leading term, or multiply by the conjugate.
• $1^{\infty}$ , $\infty^{0}$ , $0^{0}$ : set $y=f^{g}$ , take the LOGARITHM, $\ln y=g\ln f$ , which is a $0\cdot\infty$ form.
• Never forget to go back to the exponential: if $\ln y\to L$ , then $y\to e^{L}$ .

The last five forms are not harder, they just need one line of rewriting before you can differentiate.

Growth rates, to know without computing

• $\ln x\ll x^{a}\ll e^{x}$ as $x\to+\infty$ , for every $a>0$ ; for a sequence, the chain extends with $e^{n}\ll n!$ .
• $\lim_{x\to+\infty}\frac{\ln x}{x^{a}}=0$ and $\lim_{x\to+\infty}\frac{x^{a}}{e^{x}}=0$ , whatever the power.
• $\lim_{x\to 0^{+}}x\ln x=0$ : the logarithm heads to $-\infty$ , but $x$ wins.
• $\lim_{x\to+\infty}\left(1+\frac{1}{x}\right)^{x}=e$ , the most famous $1^{\infty}$ form.

\frac{\ln x}{x}

and

\frac{x^{2}}{e^{x}}

tend to

0

: the logarithm is crushed by

x

, the power by the exponential.

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.

The seven forms, and the rewriting that makes L'Hôpital's rule applicable

The rule only applies to TWO forms. The other five are rewritten first, and it is this rewriting, not the differentiation, that earns the marks.

Form met	What you write	Form obtained
$\frac{0}{0}$	nothing, the rule applies	differentiate top and bottom
Example: $\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{\cos x}{1}=1$ . At $x=0.001$ , the quotient is already $0.99999983$ .
$\frac{\infty}{\infty}$	nothing, the rule applies	differentiate top and bottom
Example: $\lim_{x\to+\infty}\frac{\ln x}{x}=\lim_{x\to+\infty}\frac{1/x}{1}=0$ . At $x=1000$ , the quotient is about $0.0069$ .
$0\cdot\infty$	$fg=\frac{f}{1/g}$	$\frac{0}{0}$ or $\frac{\infty}{\infty}$
Example: $\lim_{x\to 0^{+}}x\ln x=\lim_{x\to 0^{+}}\frac{\ln x}{1/x}=\lim_{x\to 0^{+}}(-x)=0$ . At $x=0.001$ , the product is about $-0.0069$ .
$\infty-\infty$	common denominator, or conjugate	$\frac{0}{0}$
Example: $\lim_{x\to 0}\left(\frac{1}{\sin x}-\frac{1}{x}\right)=\lim_{x\to 0}\frac{x-\sin x}{x\sin x}=0$ . At $x=0.01$ , the difference is about $0.00167$ .
$1^{\infty}$	$y=f^{g}$ , then $\ln y=g\ln f$	$0\cdot\infty$ , then $\mathrm{e}^{L}$
Example: $\lim_{x\to+\infty}\left(1+\frac{1}{x}\right)^{x}=\mathrm{e}\approx 2.718$ : $\ln y=x\ln\left(1+\frac{1}{x}\right)\to 1$ , so $y\to\mathrm{e}^{1}$ .
$\infty^{0}$	$y=f^{g}$ , then $\ln y=g\ln f$	$0\cdot\infty$ , then $\mathrm{e}^{L}$
Example: $\lim_{x\to+\infty}x^{1/x}=\mathrm{e}^{0}=1$ : $\ln y=\frac{\ln x}{x}\to 0$ . At $x=1000$ , the value is about $1.0069$ .
$0^{0}$	$y=f^{g}$ , then $\ln y=g\ln f$	$0\cdot\infty$ , then $\mathrm{e}^{L}$
Example: $\lim_{x\to 0^{+}}x^{x}=\mathrm{e}^{0}=1$ : $\ln y=x\ln x\to 0$ . At $x=0.001$ , the value is about $0.993$ .
Everything else	substitute	the answer, without differentiating
Example: $\frac{5}{0^{+}}=+\infty$ , $\frac{5}{\infty}=0$ , $\frac{0}{5}=0$ : applying L'Hôpital's rule to one of these forms gives a wrong result, and the line is counted as a method error.

The last line of the rewriting through the logarithm is the one people forget: $\ln y\to L$ concludes nothing, you must write $y\to\mathrm{e}^{L}$ . A $\ln y$ that tends to $0$ means that $y$ tends to $1$ , not to $0$ .

Growth rates, with the numbers that prove them

The hierarchy $\ln x\ll x^{a}\ll \mathrm{e}^{x}$ , extended by $\mathrm{e}^{n}\ll n!$ for a sequence, gives the answer before any computation, and a single numerical evaluation is enough to be convinced.

Quotient at $+\infty$	Which one wins	Limit
$\frac{\ln x}{x^{a}}$ , $a>0$	the power, always	$0$
Example: With $a=1$ , at $x=10^{6}$ the quotient is about $1.4\times 10^{-5}$ .
$\frac{x^{a}}{\mathrm{e}^{x}}$	the exponential, always	$0$
Example: With $a=3$ , at $x=20$ the quotient is about $1.6\times 10^{-5}$ , even though $x^{3}=8000$ looked large.
$\frac{\mathrm{e}^{n}}{n!}$	the factorial	$0$
Example: At $n=20$ , the quotient is about $2\times 10^{-10}$ : the factorial crushes even the exponential.
$x\ln x$ at $0^{+}$	the factor $x$	$0$
Example: At $x=0.001$ , the product is about $-0.0069$ : it tends to $0$ through negative values.
$\frac{\ln x}{\sqrt{x}}$	the square root, which is a power	$0$
Example: At $x=10^{6}$ , the quotient is about $0.0138$ : convergence is slow, but it does happen.

These five lines almost always save a double application of L'Hôpital's rule. On an exam, they are quoted by name: “by comparison of growth rates”, followed by the limit.

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 the quotient instead of taking the quotient of the derivatives

the whole question

What not to write

$\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{x\cos x-\sin x}{x^{2}}$

What to write

$\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{\cos x}{1}=1$

Why: L'Hôpital's rule is NOT the quotient rule: differentiate the top, differentiate the bottom, and form a quotient again. The confusion comes from the fact that both formulas deal with a fraction, but they have nothing in common, and the first one leads to an expression that is even more indeterminate.

2. Applying the rule to a determinate form

the whole question, since the answer is wrong

What not to write

$\lim_{x\to 0}\frac{\cos x}{x+1}=\lim_{x\to 0}\frac{-\sin x}{1}=0$

What to write

Substitution gives $\frac{1}{1}=1$ : the form is not indeterminate, the limit is $1$ .

Why: Outside the two allowed forms, the rule produces a number that has nothing to do with the limit. The move that protects you takes three seconds: substitute first and write down the form obtained, before differentiating anything.

3. Attacking a $0\cdot\infty$ form without rewriting it

the whole question

What not to write

“ $\lim_{x\to 0^{+}}x\ln x$ has the form $0\cdot\infty$ , I apply L'Hôpital's rule.”

What to write

“I write $x\ln x=\frac{\ln x}{1/x}$ , of the form $\frac{-\infty}{+\infty}$ , then L'Hôpital's rule gives $\frac{1/x}{-1/x^{2}}=-x\to 0$ .”

x\ln x

tends to

0

0^{+}

: the factor

x

beats

\ln x

. The form

0\cdot\infty

said nothing; the quotient

\frac{\ln x}{1/x}

shows it.

Why: The rule only knows quotients. A product must become a quotient, and there are two ways to do it: $\frac{f}{1/g}$ or $\frac{g}{1/f}$ . Choose the one that gives the simplest derivatives, here the one that puts the logarithm on top.

4. Forgetting to take the logarithm on a power form

the whole question

What not to write

“ $\left(1+\frac{1}{x}\right)^{x}\to 1^{\infty}=1$ .”

What to write

“I set $y=\left(1+\frac{1}{x}\right)^{x}$ , so $\ln y=x\ln\left(1+\frac{1}{x}\right)\to 1$ , so $y\to e$ .”

\left(1+\frac{1}{x}\right)^{x}

climbs toward

e

and stays far from

1

: a base that TENDS to

1

raised to an infinite power does not give

1

Why: $1^{\infty}$ is an INDETERMINATE form, not a computation: $1^{n}=1$ holds for a base exactly equal to $1$ , never for a base that TENDS to $1$ . Here the base equals $1+\frac{1}{x}$ , strictly greater than $1$ , and the exponent wins: the limit is $e\approx 2.718$ .

5. Forgetting to go back to the exponential

the whole question, even though the whole computation is right

What not to write

“ $\ln y\to 1$ , so the limit we want is $1$ .”

What to write

“ $\ln y\to 1$ , so $y\to e^{1}=e$ .”

Why: You have computed the limit of the LOGARITHM of the quantity you want, not the quantity itself. It is the last line, the one fatigue skips, and it turns an exact answer into a wrong one. Writing “let $y=\ldots$ ” at the start forces you to conclude about $y$ .

6. Repeating the rule forever on a limit where it goes in circles

ten minutes, then the question

What not to write

“ $\lim_{x\to+\infty}\frac{x}{\sqrt{x^{2}+1}}$ : I differentiate, it gives back a $\frac{\infty}{\infty}$ form, I start over...”

What to write

“I factor: $\frac{x}{\left|x\right|\sqrt{1+1/x^{2}}}\to 1$ at $+\infty$ .”

\frac{x}{\sqrt{x^{2}+1}}

climbs toward

1

: by dividing the top and the bottom by

x

, algebra concludes where L'Hôpital's rule goes in circles.

Why: L'Hôpital's rule can loop: here each differentiation reproduces the same form. When two successive applications simplify nothing, give up the rule and factor out the leading term, which is always possible for an algebraic function.

7. Not checking the form again between two applications

the whole question

What not to write

$\lim_{x\to 0}\frac{e^{x}-1-x}{x^{2}}=\lim\frac{e^{x}-1}{2x}=\lim\frac{e^{x}}{2}=\lim\frac{e^{x}}{0}$

What to write

After two applications, the form is $\frac{1}{2}$ , determinate: STOP, and the limit is $\frac{1}{2}$ .

Why: A third application differentiates a constant denominator and turns it into $0$ , which manufactures a fake infinite form. The rule requires checking the indeterminate form BEFORE each use, including the second and the third.

8. Applying the rule to a sequence

1 mark for rigour

What not to write

“ $u_{n}=\frac{\ln n}{n}$ , I differentiate with respect to $n$ .”

What to write

“A sequence is not differentiable. I study the FUNCTION $f(x)=\frac{\ln x}{x}$ on the real numbers, whose limit is $0$ , and deduce that of the sequence.”

The terms

u_{n}=\frac{\ln n}{n}

are the values of

f(x)=\frac{\ln x}{x}

at the integers: the limit of

f

+\infty

, computed with L'Hôpital's rule, gives that of the sequence.

Why: $n$ is an integer, and differentiation requires a continuous variable. Moving to the associated function is legitimate and is said in one sentence: if $f$ tends to $L$ at $+\infty$ and if $u_{n}=f(n)$ , then $u_{n}$ tends to $L$ . The converse, however, is false.

Which method to choose

Which rewriting for which form

Substitute first and NAME the form obtained. It, and it alone, decides what comes next.

If form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ → L'Hôpital's rule directly, differentiating top and bottom separately

Example: $\lim_{x\to 0}\frac{e^{x}-1}{x}=\lim\frac{e^{x}}{1}=1$
If form $0\cdot\infty$ → write the product as a QUOTIENT, moving one of the two factors to the denominator as its reciprocal

Example: $x\ln x=\frac{\ln x}{1/x}$
If form $\infty-\infty$ → common denominator, factoring out the leading term, or multiplying by the conjugate if there is a square root

Example: $\sqrt{x^{2}+x}-x=\frac{x}{\sqrt{x^{2}+x}+x}\to\frac{1}{2}$
If form $1^{\infty}$ , $\infty^{0}$ or $0^{0}$ → set $y$ equal to the expression, take the LOGARITHM, handle the $0\cdot\infty$ form obtained, then go back to the exponential

Example: $\ln y=x\ln\left(1+\frac{1}{x}\right)\to 1$ , so $y\to e$
If the form is not indeterminate → substitute and conclude, without any differentiation

Example: $\frac{\cos 0}{0+1}=1$

For a purely algebraic function at $\pm\infty$ , factoring out the leading term is often faster than L'Hôpital's rule, and it never loops.

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.

An indeterminate exponential form, written up for the grader

When to use it: Any limit of the form $f(x)^{g(x)}$ that gives $1^{\infty}$ , $\infty^{0}$ or $0^{0}$ .

1 Name the quantity: “Let $y=\left(1+\frac{1}{x}\right)^{x}$ .”
2 Take the logarithm: “Then $\ln y=x\ln\left(1+\frac{1}{x}\right)$ , of the form $\infty\cdot 0$ .”
3 Rewrite as a quotient: “ $\ln y=\frac{\ln\left(1+\frac{1}{x}\right)}{1/x}$ , of the form $\frac{0}{0}$ .”
4 Apply L'Hôpital's rule and compute the limit of $\ln y$ .
5 GO BACK to the exponential and conclude about $y$ .

Concluding sentence

$\ln y\to 1$ as $x\to+\infty$ , so by continuity of the exponential $y\to e^{1}=e$ .

The trap: Concluding about $\ln y$ . The question was about $y$ , and the last line, the one that goes back to the exponential, is often worth as much as the whole computation before it.

Marking: Usually 1 mark for taking the logarithm, 1 mark for rewriting as a quotient, 1 mark for L'Hôpital's rule, 1 mark for going back to the exponential.

Check before you hand in

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

Substitute before differentiating, always

Replace $x$ with the limiting value and write the form obtained in full. It is the only check that stops you from applying the rule outside its scope.

$\frac{\cos x}{x+1}$ at $0$ gives $\frac{1}{1}$ : a determinate form, no differentiation, limit $1$ .
Test numerically near the point

Evaluate the expression at a value very close to the limit point, from both sides when possible: the result must approach the announced answer.

$\left(1+\frac{1}{x}\right)^{x}$ at $x=10^{6}$ displays $2.71828$ , which confirms $e$ and rules out the answer $1$ .
Check that the form has become simpler

After one application, compare the new expression with the old one. If it is more complicated, or has the same shape, the rule will lead nowhere and you must factor.

$\frac{x}{\sqrt{x^{2}+1}}$ gives back the same form after differentiation: factoring out the leading term is the right path.
Check against the growth rates

Before any computation, identify which one wins: logarithm, power or exponential. The announced result must be consistent with this hierarchy.

$\frac{x^{100}}{e^{x}}\to 0$ at $+\infty$ , whatever the power: an infinite answer would be absurd, and a hundred applications of L'Hôpital's rule are pointless.

The typical problem, taken apart

An $\infty^{0}$ form, from the rewriting to the conclusion

Compute $\displaystyle\lim_{x\to+\infty}x^{1/x}$ .

The hierarchy of growth rates: the logarithm is crushed by every power, which is itself crushed by the exponential. It makes the answer predictable before any computation.

Step 1

Substitution: the base $x$ tends to $+\infty$ and the exponent $\frac{1}{x}$ tends to $0$ . The form is $\infty^{0}$ , INDETERMINATE.

Why

Naming the form is the first line of the solution. $\infty^{0}$ is not $1$ : the base pushes toward infinity, the exponent pushes toward $1$ , and nothing says which one wins.

Step 2

Let $y=x^{1/x}$ , then $\ln y=\frac{1}{x}\ln x=\frac{\ln x}{x}$ .

Why

The logarithm turns a power into a product, hence a power form into a $0\cdot\infty$ form. Here the rewriting as a quotient is even immediate, since the exponent is already a fraction.

Step 3

$\frac{\ln x}{x}$ has the form $\frac{\infty}{\infty}$ . L'Hôpital's rule gives $\frac{1/x}{1}=\frac{1}{x}\to 0$ .

Why

A single application is enough, and we check again that the new form, $\frac{0^{+}}{1}$ , is no longer indeterminate. This is the moment to stop, not to start over.

Step 4

So $\ln y\to 0$ , and by continuity of the exponential, $y\to e^{0}=1$ .

Why

The last line is the one that answers the question. Without it, the solution has computed the limit of the logarithm and stops halfway, which costs the mark for the conclusion.

Step 5

Numerical check: $1000^{1/1000}=e^{\ln 1000/1000}=e^{0.0069}\approx 1.0069$ , and $10^{6}$ to the power $10^{-6}$ is about $1.0000138$ ✓.

Why

Two values that come down toward $1$ confirm the answer and rule out the opposite mistake, the one that would conclude infinity because the base tends to it.

Step 6

Reading through the growth rates: $\ln y=\frac{\ln x}{x}$ , and the logarithm is crushed by any power of $x$ . So we could have announced $\ln y\to 0$ without differentiating.

Why

Knowing the hierarchy $\ln x\ll x^{a}\ll e^{x}$ lets you predict the answer, then prove it properly with L'Hôpital's rule. Predicting before computing is what protects you from sign errors.

The conclusion, written out

$\displaystyle\lim_{x\to+\infty}x^{1/x}=1$ , obtained by setting $y=x^{1/x}$ , by showing that $\ln y=\frac{\ln x}{x}\to 0$ , then by going back to the exponential.

The classic mistake on this problem: Concluding $+\infty$ “because the base tends to infinity”. In an indeterminate form, neither of the two factors decides alone: that is the whole point of the chapter, and the computation shows here that the exponent wins.

Learn by heart

• L'Hôpital's rule: $\lim\frac{f}{g}=\lim\frac{f'}{g'}$ , only for $\frac{0}{0}$ and $\frac{\infty}{\infty}$ .
• Differentiate top and bottom SEPARATELY. This is not the quotient rule.
• Check the form again before EVERY application, including the second.
• $0\cdot\infty$ is rewritten as a quotient; $\infty-\infty$ with a common denominator or the conjugate.
• $1^{\infty}$ , $\infty^{0}$ , $0^{0}$ : logarithm, then back to the exponential. The last line counts.
• $\ln x\ll x^{a}\ll e^{x}$ : the hierarchy that gives the answer before the computation.
• A sequence cannot be differentiated: go through the associated function, and say so.
• If two applications simplify nothing, factor out the leading term.

Frequently asked questions

When can L'Hôpital's rule be used?

Only on an indeterminate form zero over zero or infinity over infinity, with functions differentiable near the point and a differentiated denominator that does not vanish, and only if the limit of the quotient of the derivatives exists. On any other form, the rule gives a number that has nothing to do with the limit. So substitute first and name the form obtained.

Should you differentiate the quotient, or the top and the bottom?

The top and the bottom, separately, then form a quotient again. L'Hôpital's rule has nothing to do with the quotient rule, despite the apparent resemblance. Applying the quotient rule leads to an even more complicated expression, which is usually still indeterminate.

How do you handle a one-to-the-infinity form?

Set y equal to the expression, take the logarithm, which turns the power into a product, rewrite this product as a quotient, apply the rule, then go back to the exponential. The last step is the one people forget, and it turns an exact answer into a wrong one.

Why is one to the power infinity not equal to one?

Because the base tends to one without being equal to it. The difference, however small, is raised to a power that blows up, and the result can be anything. The classic example is one plus one over x, raised to the power x, whose limit is the number e, about 2.718.

Can L'Hôpital's rule be applied to a sequence?

Not directly, because a sequence is not differentiable, its index being an integer. Study the associated real function, compute its limit at plus infinity with the rule, then conclude that the sequence has the same limit. This step must be written out, it is worth a mark for rigour.

Practise it

Corrected exercises: L'Hôpital's rule and indeterminate forms, 201-SN3-RE (201-NYB)

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.

11 corrected exercises
110 points
150 minutes

Do the exercises

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Revision sheet: indeterminate forms and L'Hôpital's rule, 201-SN3-RE (201-NYB)

The essentials

The exact statement, and its four conditions

The seven forms, and how to rewrite them

Growth rates, to know without computing

The rules in table form

The seven forms, and the rewriting that makes L'Hôpital's rule applicable

Growth rates, with the numbers that prove them

The mistakes that cost marks

1. Differentiating the quotient instead of taking the quotient of the derivatives

2. Applying the rule to a determinate form

3. Attacking a 0⋅∞0\cdot\infty0⋅∞ form without rewriting it

4. Forgetting to take the logarithm on a power form

5. Forgetting to go back to the exponential

6. Repeating the rule forever on a limit where it goes in circles

7. Not checking the form again between two applications

8. Applying the rule to a sequence

Which method to choose

Which rewriting for which form

How the answer is expected to be written

An indeterminate exponential form, written up for the grader

Check before you hand in

The typical problem, taken apart

An ∞0\infty^{0}∞0 form, from the rewriting to the conclusion

Learn by heart

Frequently asked questions

When can L'Hôpital's rule be used?

Should you differentiate the quotient, or the top and the bottom?

How do you handle a one-to-the-infinity form?

Why is one to the power infinity not equal to one?

Can L'Hôpital's rule be applied to a sequence?

Corrected exercises: L'Hôpital's rule and indeterminate forms, 201-SN3-RE (201-NYB)

See also

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3. Attacking a $0\cdot\infty$ form without rewriting it

An $\infty^{0}$ form, from the rewriting to the conclusion