Differential Calculus 201-SN2-RE • CEGEP, Montreal

Revision sheet: L'Hôpital's rule and indeterminate forms (201-SN2-RE)

L'Hôpital's rule comes in 201-SN2-RE (formerly 201-NYA) just as the differentiation rules have been established, and the first thing you do with them is use them to compute limits that algebra cannot resolve. It is also when bad habits set in: differentiating without checking, differentiating the expression instead of the top and the bottom, or proving again, in a circle, a limit you already know.

This sheet gives what really costs marks, then the uses of the chapter within the course itself: growth rates, oblique asymptotes, differentiable extension and the accuracy of an approximation. It is written for CEGEP students in the Science program (Sciences de la nature); the corrected exercise set on the same chapter puts each of these moves to work.

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The thread of the chapter

L'Hôpital's rule is not a rule for computing limits, it is a cure for one specific pathology. Two moves follow from that: check the indeterminate form BEFORE every differentiation, because applied to a healthy limit the rule gives a result that has nothing to do with it; and ask yourself whether the limit is not, in disguise, a derivative you already know, in which case applying the rule would be circular.

This chapter is part of 201-SN2-RE, Differential Calculus

What is on this sheet

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

Topics covered

Indeterminate form
L'Hôpital's rule
Derivative at a point
Product form
Difference form
Exponential forms
Comparison of growth rates
Oblique asymptote
Continuous extension
Linear approximation
Order of contact
Change of variable

The essentials

The rule, its hypotheses, and the forms that reduce to it

• The seven indeterminate forms: $\dfrac{0}{0}$ , $\dfrac{\infty}{\infty}$ , $0\cdot\infty$ , $\infty-\infty$ , $0^{0}$ , $1^{\infty}$ , $\infty^{0}$ . Everything else, including $\dfrac{1}{0}$ and $\dfrac{\infty}{0}$ , is settled by substitution.
• RULE. If the form is $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ , if $f$ and $g$ are differentiable near $a$ with $g'$ nonzero, and if $\lim\dfrac{f'}{g'}$ EXISTS, then $\lim\dfrac{f}{g}=\lim\dfrac{f'}{g'}$ . Differentiate the top and the bottom SEPARATELY.
• Product $0\cdot\infty$ : move to the denominator the factor whose derivative COMPLICATES the expression, and keep on top the one it simplifies. The logarithm simplifies, the exponential and the root do not.
• Difference $\infty-\infty$ : common denominator, factoring out the dominant term, or the conjugate when there is a root. NEVER differentiate term by term.
• Exponential forms: set $y=f^{g}$ , compute $\lim\ln y=\lim g\ln f$ , then conclude $\lim y=e^{L}$ . The last step is forgotten one time in three.

Both limits of an oblique asymptote are almost always indeterminate: here

a=1

by substitution, then

b=\lim x\left(e^{1/x}-1\right)=1

A limit of the type $\lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}$ is a DERIVATIVE AT A POINT: it is read off the table without any computation, and applying L'Hôpital's rule to it would be circular when it is the very limit that established the derivative being used.

What the rule decides in the rest of the course

• Growth rates: $\ln x\ll x^{p}\ll b^{x}$ for all $p>0$ and $b>1$ , and between two exponentials the larger base wins. This is what tells you whether a curve comes back down to the axis or shoots off to infinity.
• Oblique asymptote: $a=\lim\dfrac{f(x)}{x}$ , finite and nonzero, then $b=\lim\left(f(x)-ax\right)$ , finite. Pushing one step further gives the SIDE from which the curve approaches its asymptote.
• Continuous extension: set $f(a)=\lim f$ . Whether the extension is differentiable is decided by a SECOND limit, that of the difference quotient, and it is not automatic.
• Accuracy of an approximation: if $\lim_{x\to 0}\dfrac{f(x)-f(0)-f'(0)x}{x^{n}}=C$ nonzero, the error of the linear approximation is approximately $Cx^{n}$ , and its SIGN tells you on which side of its tangent line the curve lies.
• Three ways the rule fails: it GOES IN CIRCLES, the quotient of derivatives has NO limit, or every round makes things WORSE. Respective remedies: algebra, squeezing, change of variable $u=\dfrac{1}{x}$ .

The exponential beats the cube: the curve rises, peaks at

x=3

, then flattens onto the axis, which is a horizontal asymptote.

The order of contact of two curves is the order of the first derivatives that stop agreeing, and $C=\dfrac{f^{(n)}(a)-g^{(n)}(a)}{n!}$ . Spotting it through the derivatives spares you any limit computation.

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. Applying the rule without checking for an indeterminate form

2 marks, even when the result happens to be right

What not to write

“ $\lim_{x\to 1}\dfrac{x^{2}+3}{x+1}=\lim\dfrac{2x}{1}=2$ .”

What to write

“At $1$ , the form is $\dfrac{4}{2}$ : determinate, so the limit is $2$ by substitution.”

Why: The rule is a cure for one specific pathology; applied to a healthy limit, it gives a number that has nothing to do with it. Here it lands on the right answer by coincidence, but on $\lim_{x\to 1}\dfrac{x^{2}+3}{x+3}$ it would give $2$ instead of $1$ . The correct move is to SUBSTITUTE first, always.

2. Differentiating the expression instead of the top and the bottom separately

3 marks, and an expression more complicated than at the start

What not to write

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

What to write

“ $\lim_{x\to 0}\dfrac{e^{x}-1}{x}=\lim\dfrac{e^{x}}{1}=1$ .”

Why: The quotient rule is used to DIFFERENTIATE a function, L'Hôpital's rule to compute a LIMIT. Both statements talk about “differentiating a quotient”, hence the confusion, but they do not produce the same expression. The alarm signal is immediate: if the result of the first step is heavier than the start, you have not applied the right rule.

3. Not checking the indeterminate form again at the second round

2 marks, and a limit declared nonexistent when it is

2

What not to write

“ $\dfrac{e^{2x}-1-2x}{x^{2}}\to\dfrac{2e^{2x}-2}{2x}\to\dfrac{2-2}{0}$ , undefined.”

What to write

“ $\dfrac{2e^{2x}-2}{2x}$ is still $\dfrac{0}{0}$ : differentiate again, $\dfrac{4e^{2x}}{2}\to 2$ .”

Why: You may apply the rule again as many times as the indeterminate form PERSISTS, provided you check it at every step. Here the student stops, believing they have hit a wall at $\dfrac{0}{0}$ , when that is precisely the green light to keep going. The expected write-up names the form before each differentiation.

4. Proving a special limit with the derivative it helped build

2 marks on a proof question, none on a computation

What not to write

“ $\lim_{x\to 0}\dfrac{\sin x}{x}=\lim\dfrac{\cos x}{1}=1$ , proved.”

What to write

“This limit is proved by a geometric squeeze; L'Hôpital's rule recovers it, it does not prove it.”

Why: The derivative of the sine was established FROM this limit. Using it to prove the limit again assumes what you want to prove. The result stays right: it is the reasoning that is circular. Same warning for $\dfrac{e^{x}-1}{x}$ and $\dfrac{\ln(1+x)}{x}$ , which ARE the derivatives that were used to build the table.

5. Forgetting to go back to the exponential

2 marks, and a result seven times too small

What not to write

“ $\ln y\to 3$ , so $\lim\left(\dfrac{x+2}{x-1}\right)^{x}=3$ .”

What to write

“ $\ln y\to 3$ , so the limit is $e^{3}\approx 20.09$ .”

\left(\frac{x+2}{x-1}\right)^{x}

approaches

e^{3}\approx 20.09

(green), not

3

(red):

3

is the limit of

\ln y

, you still have to go back to the exponential.

Why: The method computes the limit of the LOGARITHM of the expression, not that of the expression. The cure takes two lines of write-up: write “let $y=\ldots$ ” on the first line, and “therefore $y\to e^{L}$ ” on the last. The omission is all the easier because the auxiliary variable often vanishes from the paper.

6. Believing that a base tending to 1 gives a limit equal to 1

3 marks, and a misunderstanding of what an indeterminate form is

What not to write

“ $\left(1+\dfrac{2}{x}\right)^{x}\to 1$ , since the base tends to $1$ .”

What to write

“Form $1^{\infty}$ : $\ln y=x\ln\left(1+\dfrac{2}{x}\right)\to 2$ , so the limit is $e^{2}\approx 7.39$ .”

The base

1+\frac{2}{x}

tends to

1

, but the exponent runs off to infinity: the curve rises towards

e^{2}\approx 7.39

(green), never towards

1

(red). It is a

1^{\infty}

form.

Why: A base that TENDS to $1$ is not a base EQUAL to $1$ : the excess, however small, is raised to a power that tends to infinity, and the two tendencies fight it out. That is exactly what the word “indeterminate” describes. Financial translation: a rate of fifty percent compounded continuously ends the year with about ten percent more money than when compounded once a year.

7. Differentiating a difference term by term

3 marks: the rule only applies to a QUOTIENT

What not to write

“ $\lim_{x\to 1}\left(\dfrac{1}{\ln x}-\dfrac{1}{x-1}\right)$ : I differentiate each term.”

What to write

“Common denominator first: $\dfrac{(x-1)-\ln x}{(x-1)\ln x}$ , then the rule twice, limit $\dfrac{1}{2}$ .”

Why: L'Hôpital's rule is about $\dfrac{f}{g}$ , never about a sum or a difference. An $\infty-\infty$ form must therefore be rewritten as a quotient BEFORE any differentiation, by a common denominator, by factoring out the dominant term, or with the conjugate. Faced with a difference containing a root, always try the conjugate first.

8. Insisting when the rule fails

3 marks, and a limit wrongly declared nonexistent

What not to write

“ $\lim_{x\to\infty}\dfrac{2x+\cos x}{3x+\sin x}$ : the quotient of derivatives oscillates, so the limit does not exist.”

What to write

“The rule does not conclude, but dividing by $x$ the limit is $\dfrac{2}{3}$ .”

The quotient (solid line) settles on

\frac{2}{3}

, while the quotient of the derivatives (red) oscillates endlessly. The rule does not conclude; dividing by

x

does.

Why: The rule is a ONE-WAY implication: it states something IF the quotient of derivatives has a limit, and says nothing at all otherwise. There are three ways of failing, going in circles, the quotient of derivatives with no limit and getting worse, with their remedies: algebra, squeezing, change of variable. The alarm signal: if after two rounds the expression is not simpler, it is not the right tool.

Which method to choose

Which tool, depending on the form obtained by substitution

The result of substituting into the expression

If a determinate form, even $\dfrac{1}{0}$ or $\dfrac{\infty}{0}$ → conclude directly, without differentiating anything

Example: $\lim_{x\to 0^{+}}\dfrac{\ln x}{x}=-\infty$
If a quotient of polynomials or a difference of roots → ALGEBRA first: factor out the dominant term, or multiply by the conjugate

Example: $\lim_{x\to\infty}\left(x-\sqrt{x^{2}-4x}\right)=2$
If a difference quotient, $\dfrac{f(x)-f(a)}{x-a}$ → it is $f'(a)$ : read the table, do not differentiate the quotient

Example: $\lim_{x\to 2}\dfrac{x^{5}-32}{x-2}=5\times 2^{4}=80$
If a genuine $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ → L'HÔPITAL'S RULE, checking the form again before every round

Example: $\lim_{x\to 0}\dfrac{e^{2x}-1-2x}{x^{2}}=2$ , two rounds
If $0\cdot\infty$ or $\infty-\infty$ → rewrite as a QUOTIENT, then apply the rule

Example: $\lim_{x\to\infty}x\left(\dfrac{\pi}{2}-\arctan x\right)=1$
If $0^{0}$ , $1^{\infty}$ or $\infty^{0}$ → LOGARITHM: $\ln y=g\ln f$ , then $y\to e^{L}$

Example: $\lim_{x\to\infty}\left(1+\dfrac{2}{x}\right)^{x}=e^{2}$

Two growth-rate limits spare you any computation: $\lim\dfrac{(\ln x)^{n}}{x^{p}}=0$ and $\lim\dfrac{x^{p}}{b^{x}}=0$ for $p>0$ and $b>1$ . Recognizing them saves several rounds of the rule on the exam.

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 the computation of an indeterminate limit

When to use it: The question asks for a limit and substitution does not conclude.

1 SUBSTITUTE and name the form obtained, in words: “form $\dfrac{0}{0}$ ”, “form $1^{\infty}$ ”. Without this line, what follows has no justification.
2 Choose the tool and say so: algebra, derivative at a point, rewriting as a quotient, logarithm, or L'Hôpital's rule.
3 At each application of the rule, name the form again BEFORE differentiating, and stop as soon as it is no longer indeterminate.
4 Conclude with a sentence that gives the limit, and for an exponential form go back to the exponential explicitly.

Concluding sentence

“At $0$ , the quotient has the form $\dfrac{0}{0}$ ; L'Hôpital's rule gives $\dfrac{2e^{2x}-2}{2x}$ , still of the form $\dfrac{0}{0}$ ; a second application gives $\dfrac{4e^{2x}}{2}$ , which tends to $2$ . Therefore the limit is $2$ .”

The trap: Writing a chain of equalities without ever naming the forms. The marker cannot check that the rule was applicable, and takes off the justification marks even when the final result is right.

Marking: Usually 1 mark for identifying the form, 2 for applications of the rule that are properly justified, 1 for the computation, 1 for the conclusion.

Check before you hand in

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

Substitution first, always

Before pulling out the rule, replace $x$ by $a$ and look. A determinate form is settled at once, and applying L'Hôpital's rule to a healthy limit gives a wrong result.

$\dfrac{x^{2}+3}{x+1}$ at $1$ equals $\dfrac{4}{2}$ : nothing to differentiate.
The quotient of derivatives must be SIMPLER

If the expression obtained after a differentiation is heavier than at the start, either you applied the quotient rule by mistake, or L'Hôpital's rule is not the right tool.

$\dfrac{e^{-1/x}}{x}$ becomes $\dfrac{e^{-1/x}}{x^{2}}$ : change method.
The order of magnitude of the result

A $1^{\infty}$ form almost always gives an exponential, never $1$ . A growth-rate comparison always gives $0$ or $\infty$ , never an in-between number. A result that contradicts these landmarks must be redone.

$\left(1+\dfrac{2}{x}\right)^{x}\to e^{2}\approx 7.39$ , not $1$ .
The thirty-second numerical test

When you announce a limit at $0$ , evaluate the expression at $x=0.01$ ; when you announce a limit at infinity, evaluate it at $x=1000$ . Agreement confirms, disagreement reveals a lost sign or factor.

$\dfrac{e^{0.1}-1}{0.1}=1.0517$ , consistent with the limit $1$ and the slope $\dfrac{1}{2}$ .
The sign of the gap to the tangent line

When the limit is used to compare a curve with its tangent line, the SIGN of the result tells you on which side the curve lies, and an odd power signals a crossing, hence an inflection.

$\dfrac{\arctan x-x}{x^{3}}\to-\dfrac{1}{3}$ : inflection at $0$ .

The typical problem, taken apart

A complete oblique asymptote, from the slope to the side

Find the oblique asymptote of $f(x)=x\,e^{1/x}$ at $+\infty$ , then say from which side the curve approaches it.

Far from the origin, the graph of

x\,e^{1/x}

seems to run along an oblique line: the example finds which one, with two limits, then from which side the curve approaches it.

Step 1

$a=\lim_{x\to\infty}\dfrac{f(x)}{x}=\lim e^{1/x}=e^{0}=1$ .

Why

Substitute first: $\dfrac{1}{x}\to 0$ , so there is no indeterminate form. Pulling out the rule here would be time wasted.

Step 2

$b=\lim_{x\to\infty}\left(x\,e^{1/x}-x\right)=\lim x\left(e^{1/x}-1\right)$ , form $\infty\cdot 0$ .

Why

An oblique asymptote calls for TWO limits, and the second one is almost always indeterminate. Forgetting to compute it gives a line with the right slope but in the wrong place.

Step 3

Set $u=\dfrac{1}{x}\to 0^{+}$ : $b=\lim_{u\to 0}\dfrac{e^{u}-1}{u}=1$ .

Why

The change of variable turns a limit at infinity into a limit at $0$ , where the form is easier to read. And you recognize here the derivative of the exponential at $0$ : no differentiation is needed.

Step 4

The asymptote is therefore $y=x+1$ .

Why

Write the whole line. Answering “slope $1$ ” without the y-intercept is worth half the marks.

Step 5

Gap: $f(x)-(x+1)=\dfrac{e^{u}-1-u}{u}\approx\dfrac{u}{2}=\dfrac{1}{2x}$ , positive.

Why

Push one step further, using $\dfrac{e^{u}-1-u}{u^{2}}\to\dfrac{1}{2}$ . That is what gives the SIDE, information that a wrong sketch loses.

The conclusion, written out

“The line $y=x+1$ is an oblique asymptote to the graph of $f$ at $+\infty$ , and the curve lies ABOVE its asymptote, the gap being approximately $\dfrac{1}{2x}$ , still $0.005$ at $x=100$ .”

The classic mistake on this problem: Stopping at the slope without computing the y-intercept, or finding the asymptote without saying from which side the curve approaches it.

Learn by heart

• Seven indeterminate forms; everything else is settled by substitution.
• Differentiate the top and the bottom SEPARATELY: this is not the quotient rule.
• Name the form again BEFORE every new differentiation.
• A difference quotient is a derivative: read it, do not differentiate it.
• Products and differences are rewritten as a quotient before any differentiation.
• Exponential form: logarithm, then BACK to the exponential.
• A base that tends to $1$ is not a base equal to $1$ .
• $\ln x\ll x^{p}\ll b^{x}$ : the hierarchy spares you the computation.
• Oblique asymptote: two limits, then one step further for the side.
• Three possible failures: going in circles, no limit, getting worse.

Frequently asked questions

When are you allowed to apply L'Hôpital's rule?

Only if substitution gives the form zero over zero or infinity over infinity, if both functions are differentiable near the point, if the derivative of the denominator is not zero there, and if the quotient of the derivatives does have a limit. These four conditions are not decorative: applied to a limit that was not indeterminate, the rule gives a result that has nothing to do with it, and when the quotient of the derivatives has no limit, it says nothing at all, which certainly does not mean that the original limit does not exist.

Why can't you prove the limit of sine x over x with L'Hôpital's rule?

Because it would be circular. The derivative of the sine was established from this very limit, proved by a geometric squeeze in the chapter on limits. Using the derivative to prove the limit again amounts to assuming what you want to prove. The result stays correct, it is the proof that is invalid, and a marker takes off the marks on a proof question while granting them on a plain computation. The same warning applies to the limit of e to the x minus one over x and to that of the logarithm of one plus x over x.

How do you handle a form like zero to the power zero or one to the power infinity?

Always through the logarithm, in three steps. Set y equal to the expression, write the logarithm of y as the product of the exponent by the logarithm of the base, which gives a zero times infinity form you already know, and rewrite it as a quotient. Compute the limit of this logarithm, then, and this is the step forgotten one time in three, go back to the exponential. If the limit of the logarithm is three, the limit of the expression is e to the power three, about twenty, and not three.

What should you do when the rule goes in circles?

Change tools, because insisting will get you nowhere. Three cases come up. If the quotient of derivatives gives back the original quotient, as for a square root divided by a polynomial, factor out the dominant term at the top and the bottom. If the quotient of derivatives oscillates without converging, simplify the expression or squeeze the bounded terms. And if every differentiation makes the expression heavier, set u equal to one over x, which turns a limit at zero into a limit at infinity or the other way round. The alarm signal is simple: after two rounds, the expression should be simpler than at the start.

What is this chapter used for in the rest of the course?

Four things, all used from the chapter on curve sketching on. First, comparing growth rates, which tells you whether a curve comes back down to the axis or shoots off to infinity. Next, finding oblique asymptotes, whose two limits are almost always indeterminate. Then, extending a function by continuity and deciding whether the extension is differentiable, which takes a second limit. Finally, putting a number on the accuracy of a linear approximation, by reading in a limit the first neglected term and its sign.

Practise it

Corrected exercises: L'Hôpital's rule and indeterminate forms, 201-SN2-RE

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.

16 corrected exercises
160 points
225 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc1-lhopital-indeterminate-forms. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: L'Hôpital's rule and indeterminate forms (201-SN2-RE)

The essentials

The rule, its hypotheses, and the forms that reduce to it

What the rule decides in the rest of the course

The mistakes that cost marks

1. Applying the rule without checking for an indeterminate form

2. Differentiating the expression instead of the top and the bottom separately

3. Not checking the indeterminate form again at the second round

4. Proving a special limit with the derivative it helped build

5. Forgetting to go back to the exponential

6. Believing that a base tending to 1 gives a limit equal to 1

7. Differentiating a difference term by term

8. Insisting when the rule fails

Which method to choose

Which tool, depending on the form obtained by substitution

How the answer is expected to be written

Writing up the computation of an indeterminate limit

Check before you hand in

The typical problem, taken apart

A complete oblique asymptote, from the slope to the side

Learn by heart

Frequently asked questions

When are you allowed to apply L'Hôpital's rule?

Why can't you prove the limit of sine x over x with L'Hôpital's rule?

How do you handle a form like zero to the power zero or one to the power infinity?

What should you do when the rule goes in circles?

What is this chapter used for in the rest of the course?

Corrected exercises: L'Hôpital's rule and indeterminate forms, 201-SN2-RE

See also

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