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L'Hopital's Rule: A Mathematical Theorem for Limits, Lecture notes of Pre-Calculus

An introduction to L'Hopital's Rule, a theorem used to evaluate limits of functions. the statement of the theorem, a proof of the special case when the limit is 0/0 and a finite point, and examples of how to apply L'Hopital's Rule to various limits. The document also includes a list of indeterminate forms and their corresponding limits.

What you will learn

  • What are the indeterminate forms and how are they related to L'Hopital's Rule?
  • What is L'Hopital's Rule?
  • How is L'Hopital's Rule used to evaluate limits?
  • How is L'Hopital's Rule proven?
  • What is the statement of L'Hopital's Rule?

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2017-18 MATH1010
Lecture 15: L’Hˆopital’s Rule
Charles Li
1 L’Hˆopital’s Rule
(Source: mooculus textbook)
Derivatives allow us to take problems that were once difficult to
solve and convert them to problems that are easier to solve. Let us
consider l’Hˆopital’s rule:
Theorem 1.1 (L’Hˆopital’s Rule).Let f(x)and g(x)be functions
that are differentiable near a. If
lim
xaf(x) = lim
xag(x) = 0 or ± ,
and limxaf0(x)
g0(x)exists, and g0(x)6= 0 for al l xnear a, then
lim
xa
f(x)
g(x)= lim
xa
f0(x)
g0(x).
Remark
1. L’Hˆopital’s rule applies even when limxaf(x) = ±∞ and
limxag(x) = ±∞.
2. acan be +and −∞.
This theorem is somewhat difficult to prove, in part because it in-
corporates so many different possibilities, we will prove the special
case when ais finite.
Proof. Here we proof the special type 0/0 and afinite.
By Cauchy’s mean value theorem, there exists ξbetween aand x
such that f(x)
g(x)=f(x)f(a)
g(x)g(a)=f0(ξ)
g0(ξ).
When xa,xi (which depends on x) also tends to a. So
lim
xa
f(x)
g(x)= lim
ξa
f0(ξ)
g0(ξ).
1
pf3
pf4
pf5
pf8
pf9
pfa

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2017-18 MATH

Lecture 15: L’Hˆopital’s Rule Charles Li

1 L’Hˆopital’s Rule

(Source: mooculus textbook) Derivatives allow us to take problems that were once difficult to solve and convert them to problems that are easier to solve. Let us consider l’Hˆopital’s rule:

Theorem 1.1 (L’Hˆopital’s Rule). Let f (x) and g(x) be functions that are differentiable near a. If

lim x→a f (x) = lim x→a g(x) = 0 or ± ∞,

and limx→a f^

′(x) g′(x) exists, and^ g

′(x) 6 = 0 for all x near a, then

lim x→a

f (x) g(x)

= lim x→a

f ′(x) g′(x)

Remark

  1. L’Hˆopital’s rule applies even when limx→a f (x) = ±∞ and limx→a g(x) = ±∞.
  2. a can be +∞ and −∞. This theorem is somewhat difficult to prove, in part because it in- corporates so many different possibilities, we will prove the special case when a is finite.

Proof. Here we proof the special type 0/0 and a finite. By Cauchy’s mean value theorem, there exists ξ between a and x such that f (x) g(x)

f (x) − f (a) g(x) − g(a)

f ′(ξ) g′(ξ)

When x → a, xi (which depends on x) also tends to a. So

lim x→a

f (x) g(x)

= lim ξ→a

f ′(ξ) g′(ξ)

Next we prove the special case ∞/∞ and a finite (can be skipped). Suppose lim x→a f (x) = ∞, lim x→a g(x) = ∞.

Let b be a number very closed to a, such that g′(x) is nonzero for x between a and b. Then by Cauchy’s mean value theorem, there exists ξ, between a and b, such that f (x) − f (b) g(x) − g(b)

f ′(ξ) g′(ξ)

Then we have f (x) g(x)

f ′(ξ) g′(ξ)

1 − g g((xb)) 1 − (^) ff ((bx))

Because lim ξ→ 0

f ′(ξ) g′(ξ)

exists. Also lim x→∞

g(b) g(x)

= 0 = lim x→∞

f (b) f (x)

. Hence

lim x→a

1 − (^) gg((xb)) 1 − (^) ff ((xb))

So lim x→a

f (x) g(x)

= lim ξ→a

f ′(ξ) g′(ξ)

L’Hˆopital’s rule allows us to investigate limits of indeterminate form.

Definition 1.1. 0/0 This refers to a limit of the form limx→a f g^ ((xx)) where f (x) → 0 and g(x) → 0 as x → a.

∞∞∞/∞∞∞ This refers to a limit of the form limx→a f g^ ((xx)) where f (x) → ∞

and g(x) → ∞ as x → a. 0 ·∞·∞·∞ This refers to a limit of the form limx→a (f (x) · g(x)) where f (x) → 0 and g(x) → ∞ as x → a.

∞∞∞–∞∞∞ This refers to a limit of the form limx→a (f (x) − g(x)) where f (x) → ∞ and g(x) → ∞ as x → a. 1 ∞∞∞^ This refers to a limit of the form limx→a f (x)g(x)^ where f (x) → 1 and g(x) → ∞ as x → a.

Again set f (x) = sin x, g(x) = 2x. Then f ′(x) = cos x and g′(x) =

lim x→ 0 f (x) = sin 0 = 0, lim x→ 0 g(x) = 2x = 0.

So by L’Hˆopital’s rule

lim x→ 0

sin x 2 x

= lim x→ 0

cos x 2

Hence

lim x→ 0

1 − cos x x^2

Example 1.3 (0 00 /// 00 0). Compute

lim x→ 0

2 x^ − 1 x

Answer. Let f (x) = 2x^ and g(x) = x, then by by L’Hˆopital’s rule

lim x→ 0

2 x^ − 1 x

= lim x→ 0

f ′(x) g′(x)

= lim x→ 0

(ln 2)2x 1

= ln 2.

 Our next set of examples will run through the remaining inde- terminate forms one is likely to encounter.

Example 1.4 (∞∞∞/∞∞∞). Compute

lim x→π/2+

sec(x) tan(x)

Answer. Set f (x) = sec(x) and g(x) = tan(x). Both f (x) and g(x) are differentiable near π/2. Additionally,

lim x→π/2+

f (x) = lim x→π/2+

g(x) = −∞.

This situation is ripe for l’Hˆopital’s Rule. Now

f ′(x) = sec(x) tan(x) and g′(x) = sec^2 (x).

L’Hˆopital’s rule tells us that

lim x→π/2+

sec(x) tan(x)

= lim x→π/2+

sec(x) tan(x) sec^2 (x)

= lim x→π/2+

sin(x) = 1.

Example 1.5 (∞∞∞/∞∞∞). Compute

lim x→+∞

ln(ex^ + 1) ln(e^2 x^ + 1)

Answer. Let f (x) = ln(ex^ + 1) and g(x) = ln(e^2 x^ + 1). Then limx→+∞ f (x) = +∞ and limx→+∞ g(x) = +∞. By l’Hˆopital’s Rule

lim x→+∞

f (x) g(x)

= lim x→+∞

f ′(x) g′(x)

lim x→+∞

ex ex+ 2 e^2 x e^2 x+

= lim x→+∞

e^2 x^ + 1 2 ex(ex^ + 1)

= lim x→+∞

1 + e−^2 x 2(1 + e−x)

Example 1.6 (0 ·∞·∞·∞). Compute

lim x→0+ x ln x.



Answer. This doesn’t appear to be suitable for l’Hˆopital’s Rule. As x approaches zero, ln x goes to −∞, so the product looks like

(something very small) · (something very large and negative).

This product could be anything—a careful analysis is required. Write

x ln x =

ln x x−^1

Answer. Here we simply need to write each term as a fraction,

lim x→ 0 (cot(x) − csc(x)) = lim x→ 0

cos(x) sin(x)

sin(x)

= lim x→ 0

cos(x) − 1 sin(x)

Setting f (x) = cos(x) − 1 and g(x) = sin(x), both functions are differentiable near zero and

lim x→ 0 (cos(x) − 1) = lim x→ 0 sin(x) = 0.

We may now apply l’Hˆopital’s rule. Write

f ′(x) = − sin(x) and g′(x) = cos(x),

so

lim x→ 0 (cot(x) − csc(x)) = lim x→ 0

cos(x) − 1 sin(x)

= lim x→ 0

− sin(x) cos(x)

Sometimes one must be slightly more clever.

Example 1.9 (∞∞∞–∞∞∞). Compute

lim x→∞

x^2 + x − x

Answer. Again, this doesn’t appear to be suitable for l’Hˆopital’s Rule. A bit of algebraic manipulation will help. Write

lim x→∞

x^2 + x − x

= lim x→∞

x

1 + 1/x − 1

= lim x→∞

1 + 1/x − 1 x−^1

Now set f (x) =

1 + 1/x − 1, g(x) = x−^1. Since both functions are differentiable for large values of x and

lim x→∞

1 + 1/x − 1) = lim x→∞

x−^1 = 0,

we may apply l’Hˆopital’s rule. Write

f ′(x) = (1/2)(1 + 1/x)−^1 /^2 · (−x−^2 ) and g′(x) = −x−^2

so

lim x→∞

x^2 + x − x

= lim x→∞

1 + 1/x − 1 x−^1

= lim x→∞

(1/2)(1 + 1/x)−^1 /^2 · (−x−^2 ) −x−^2 = lim x→∞

1 + 1/x

=

Exponential Indeterminate Forms

There is a standard trick for dealing with the indeterminate forms

1 ∞, 00 , ∞^0.

Given u(x) and v(x) such that

lim x→a u(x)v(x)

falls into one of the categories described above, rewrite as

lim x→a ev(x) ln(u(x))

and then examine the limit of the exponent

lim x→a v(x) ln(u(x)) = lim x→a

ln(u(x)) v(x)−^1

using l’Hˆopital’s rule. Since these forms are all very similar, we will only give a single example.

Example 1.10 (1∞∞∞). Compute

lim x→∞

x

)x .

So now look at the limit of the exponent

lim x→ 0 +

(ln x)(sin x) = lim x→ 0 +

ln x 1 sin x

Let f (x) = ln x, g(x) = (^) sin^1 x. Apply l’Hˆopital’s rule, the limit is

lim x→ 0 +

f ′(x) g′(x)

= lim x→ 0 +

1 /x cos x sin^2 x

= lim x→ 0 +^

sin^2 x x cos x

Apply l’Hˆopital’s rule again, the above is

= lim x→ 0 +^

2 sin x cos x cos x − x sin x

Hence lim x→ 0 +^

xsin^ x^ = e^0 = 1.



Example 1.12 (∞ ∞ ∞^000 ).

lim x→+∞

(ex^ + x)

1 x (^).

Answer.

lim x→+∞ (ex^ + x)

1 x (^) = lim x→+∞ e

ln(ex+x) x (^).

So now look at the limit of the exponent and apply l’Hˆopital’s rule, the limit is

lim x→+∞

ln(ex^ + x) x

= lim x→+∞

ex ex^ + 1

Apply l’Hˆopital’s rule again, the limit is

= lim x→+∞

ex ex^

Hence lim x→+∞ (ex^ + x)

(^1) x = e^1 = e.



Exercise 1.1. limx→ 0 cossin^ x −x 1 answer. 0

Exercise 1.2. limx→∞ e x x^3 answer.^ ∞

Exercise 1.3. limx→∞

x^2 + x −

x^2 − x answer. 1

Exercise 1.4. limx→∞ lnx^ x answer. 0

Exercise 1.5. limx→∞ ln√^ xx answer. 0

Exercise 1.6. limx→∞ e

x+e−x ex−e−x^ answer.^1

Exercise 1.7. limx→ 0

√9+x− 3 x answer.^1 /^6

Exercise 1.8. limt→1+ (^) t(1 (^2) −/t 2 )t−+1^1 answer. −∞

Exercise 1.9. limx→ 2 2 −

√x+ 4 −x^2 answer.^1 /^16

Exercise 1.10. limt→∞ t+5−^2 /t−^1 /t

3 3 t+12− 1 /t^2 answer.^1 /^3

Exercise 1.11. limy→∞

√y+1+√y− 1 y answer.^0

Exercise 1.12. limx→ 1

√x− 1 √ (^3) x− 1 answer. 3 / 2

Exercise 1.13. limx→ 0 (1−x)

1 / (^4) − 1 x answer.^ −^1 /^4 Exercise 1.14. limt→ 0

t + (^1) t

((4 − t)^3 /^2 − 8) answer. − 3

Exercise 1.15. limt→0+

1 t +^ √^1 t

t + 1 − 1) answer. 1 / 2

Exercise 1.16. limx→ 0 x √^2 2 x+1− 1 answer.^0

Exercise 1.17. limu→ 1 (u−1)

3 (1/u)−u^2 +3/u− 3 answer.^0

Exercise 1.18. limx→ 0 2+(1 3 −(2/x/x)) answer. − 1 / 2

Exercise 1.19. limx→0+ 1+5/

√x 2+1/√x answer.^5

Exercise 1.20. limx→0+ 3+x − 1 / (^2) +x− 1 2+4x−^1 /^2 answer.^ ∞

Exercise 1.21. limx→∞ x+x

1 / (^2) +x 1 / 3 x^2 /^3 +x^1 /^4 answer.^ ∞

Exercise 1.22. limt→∞ 1 −

√ (^) t t+ 2 −

√ (^4) t+ t+

answer. 2 / 7

Exercise 1.23. limt→∞

1 − (^) t−t 1 1 −

√ (^) t t− 1

answer. 2

Exercise 1.24. limx→−∞ x+x − 1 1+√ 1 −x answer.^ −∞