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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
- L’Hˆopital’s rule applies even when limx→a f (x) = ±∞ and limx→a g(x) = ±∞.
- 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.^ −∞
