Chapter 8: Taylor’s theorem and L’Hospital’s rulekab/252/ch8.pdf · Chapter 8: Taylor’s theorem and L’Hospital’s rule Theorem: [Inverse Mapping Theorem] Suppose that a

Chapter 8: Taylor’s theorem and L’Hospital’s rule

Theorem: [Inverse Mapping Theorem] Suppose that a < b and f : [a, b] → R. Giventhat f ′(x) > 0 for all x ∈ (a, b) then f−1 is differentiable on (f(a), f(b)) and

(f−1)′ = 1/(f ′ ◦ f−1).

Proof: 1. Let y0 ∈ (f(a), f(b)). Since f is strictly increasing there is an x0 ∈ (a, b) withf(x0) = y0. This is so if and only if f−1(y0) = x0.

2. Define a function H : [a, b] → R by,

H(x) :=

x− x0

f(x)− f(x0)x 6= x0;

1

f ′(f−1(y0))x = x0.

Figure 23: Motivation for the definition of H.

3. Then

limx→x0

H(x) = 1/f ′(x0) = H(x0) = H(f−1(y0)) and so H is continuous at x0 = f−1(y0).

4. Moreover from the earlier work on continuity for monotonic functions we know thatf−1 is continuous on [f(a), f(b)]. Therefore,

1/f ′(x0) = H(x0) = H(f−1(y0)) = H

(limy→y0

f−1(y)

)= lim

y→y0

H(f−1(y))

= limy→y0

f−1(y)− f−1(y0)

y − y0

= (f−1)′(y0) 2

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Example: f(x) = x2 + 1 on (0, 1) has inverse f−1(y) =√

y − 1 on (1, 2) and

(f−1)′(y) =1

2√

y − 1=

1

2x.

Remark: The hypothesis that f ′(x) > 0 for all x ∈ (a, b) is essential. In fact if f isstrictly increasing and differentiable on (a, b) but f ′(x0) = 0 for some x0 ∈ (a, b) then theinverse, f−1 is not differentiable at f(x0). Indeed, if f−1 were differentiable at f(x0) thenby the chain rule we would have, 1 = (f−1 ◦ f)′(x0) = (f−1)′(f(x0)) · f ′(x0) = 0; whichis impossible. Therefore, f−1 cannot be differentiable at f(x0). The function f(x) := x3

is strictly increasing and differentiable on R however x 7→ 3√

x (the inverse of f) is notdifferentiable at f(0) = 0.

Figure 24: Differentiable function with a non-differentiable inverse.

Theorem: [Cauchy’s Mean Value Theorem]. Suppose that a < b and f : [a, b] → R andg : [a, b] → R are continuous on [a, b] and differentiable on (a, b). If g′(x) 6= 0 for allx ∈ (a, b) then there exists a point x0 ∈ (a, b) such that

f(b)− f(a)

g(b)− g(a)=

f ′(x0)

g′(x0).

Proof: Consider the auxiliary function h : [a, b] → R defined by the 3× 3 determinant,

h(x) :=

∣∣∣∣∣∣f(x) g(x) 1f(a) g(a) 1f(b) g(b) 1

∣∣∣∣∣∣ .

Now h is continuous on [a, b] and differentiable on (a, b) and h(a) = h(b) = 0 so by Rolle’stheorem there exists a point x0 ∈ (a, b) such that h′(x0) = 0; that is,∣∣∣∣∣∣

f ′(x0) g′(x0) 0f(a) g(a) 1f(b) g(b) 1

∣∣∣∣∣∣ = 0.

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ie: f ′(x0) · [g(b) − g(a)] = g′(x0) · [f(b) − f(a)]. Now since g′(x) 6= 0 for all x ∈ (a, b),Rolle’s theorem tells us that g(b)− g(a) 6= 0 and so the result follows. 2

Remark: Cauchy’s mean value theorem has a geometric interpretation. If we considerthe curve defined by the parametric equation α(t) := (f(t), g(t)), t ∈ [a, b]. Then theconclusion of the theorem is that there exists a point (f(x0), g(x0)) on the curve such thatthe slope, g′(x0)/f

′(x0) of the tangent line to the curve at that point is equal to the slopeof the line segment joining the end points of the curve.

Figure 25: Cauchy’s Mean Value Theorem interpreted.

The next theorem has sometimes been said to be the most important in Calculus orAnalysis. We use the notation for higher derivatives, f (0)(x) = f(x), f (1)(x) = f ′(x) and,in general for n ∈ N, f (n+1)(x) = (f (n))′(x). If f (n) is differentiable on an interval, thenf (n+1) exists on the interval.

Theorem: [Taylor’s Theorem] Suppose that a < b and f : [a, b] → R. If f (n) is continuouson [a, b] and differentiable on (a, b) then for each x ∈ (a, b] there exists a point ζ ∈ (a, x)such that f(x) = Pn(x) + Rn(x), where

Pn(x) := f(a) +n∑

k=1

f (k)(a)(x− a)k

k!and Rn(x) := f (n+1)(ζ)

(x− a)n+1

(n + 1)!.

Note: The conclusion of the Mean Value Theorem can be written

f(x) = f(a) + f (1)(ζ)(x− a)1 = P0(x) + R0(x),

so can be regarded as Taylor’s Theorem of order 0.

Proof: 1. Fix x0 ∈ (a, b] and define M ∈ R by,

f(x0) = f(a) +n∑

k=1

f (k)(a)(x0 − a)k

k!+ M · (x0 − a)n+1

(n + 1)!

We need to show that there exists a ζ ∈ (a, x0) such that f (n+1)(ζ) = M .

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2. Consider the auxiliary function g : [a, x0] → R defined by,

g(x) := −f(x0) + f(x) +n∑

k=1

f (k)(x)(x0 − x)k

k!+ M · (x0 − x)n+1

(n + 1)!.

3. Now g is continuous on [a, x0] and differentiable on (a, x0) and g(a) = g(x0) = 0.Therefore, by Rolle’s theorem there exists a point ζ ∈ (a, x0) such that g′(ζ) = 0.

4. But

g′(x) = f ′(x) +n∑

k=1

{f (k+1)(x)

(x0 − x)k

k!− f (k)(x)

(x0 − x)k−1

(k − 1)!

}−M · (x0 − x)n

n!

=(x0 − x)n

n!· {f (n+1)(x)−M} for all x ∈ (a, x0).

Therefore, f (n+1)(ζ) = M . This completes the proof. 2

In Taylor’s theorem the polynomial Pn(x) is called the n-th degree Taylor polynomial forf at a and Rn(x) is called the Lagrange remainder.

Suppose that a < b and f : [a, b] → R. If for each fixed x ∈ [a, b],

limn→∞

Rn(x) = 0.

Then for each x ∈ [a, b], the series of powers (a so-called power series),

∞∑n=0

f (n)(a)(x− a)n

n!

converges to f(x).

Functions with this property are common, e.g. ex, sin(x), polynomials, rational functions.They are called real analytic.

L’Hospital’s Theorems

Recall that if limx→x0

f(x) = L1 and limx→x0

g(x) = L2, then

limx→x0

f(x)

g(x)=

L1

L2

, provided L2 6= 0.

We now look into the case when L1 = L2 = 0 (there’s no point in considering the caseL1 6= 0 and L2 = 0 since the limit will be ±∞).

Theorem: Suppose that a < b and f : [a, b] → R and g : [a, b] → R. If f(x0) = g(x0) = 0,g′(x0) 6= 0 and both f ′(x0) and g′(x0) exist at some point x0 ∈ (a, b), then

limx→x0

f(x)

g(x)=

f ′(x0)

g′(x0).

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Figure 26: Convergence of Taylor polynomials to a real analytic function.

Proof: The trick in this proof is to multiply and divide by (x− x0).

limx→x0

f(x)

g(x)= lim

x→x0

f(x)− f(x0)

g(x)− g(x0)

= limx→x0

(f(x)− f(x0)

x− x0

)·(

x− x0

g(x)− g(x0)

)= f ′(x0)/g

′(x0). 2

Theorem: Suppose that a < b and f : [a, b] → R and g : [a, b] → R are continuous on[a, b]. Let x0 be any point in (a, b) such that f(x0) = g(x0) = 0 and g′(x) 6= 0 for allx 6= x0. Then,

limx→x0

f(x)

g(x)= lim

x→x0

f ′(x)

g′(x), whenever the limit on the right exists

Proof: By Cauchy’s mean value theorem there exists for each x ∈ (a, b) a point ζx

between x0 and x such that

f(x)

g(x)=

f(x)− f(x0)

g(x)− g(x0)=

f ′(ζx)

g′(ζx).

Then

limx→x0

f(x)

g(x)= lim

x→x0

f(x)− f(x0)

g(x)− g(x0)= lim

x→x0

f ′(x)

g′(x),

because when x tends to x0, ζx tends to x0 . 2

Notes: 1. The previous theorem is also true when limx→x0

f(x) = ±∞ and limx→x0

g(x) = ±∞and may also be extended to the case when x0 is replaced by ±∞.

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2. It can also be iterated by replacing f ′ by f ′′ and g′ by g′′ e.t.c for higher orderderivatives. If

f(x0) = f ′(x0) = g(x0) = g′(x0)

and the limit of the ratio of the second derivatives exists then

limx→x0

f(x)

g(x)= lim

x→x0

f ′(x)

g′(x)= lim

x→x0

f ′′(x)

g′′(x).

Exercises

1. Calculate the derivative of the function f : R → R defined by, f(x) := x3.

2. Consider the function f : (0,∞) → R defined by, f(x) := loge(x). Show that theinverse of f exists and is differentiable. Moreover, show that (f−1)′(x) = f−1(x) for allx ∈ R. Note: the function f−1 is usually called the exponential function.

3. Let f : (0,∞) → R be defined by, f(x) := x · loge(1 + 1/x).

(a) Calculate limx→∞ f(x). Hint: Consider the derivative of the function g : (0,∞) → Rdefined by, g(x) := loge(x) at x = 1.

(b) Show that limn→∞

n · loge(1 + 1/n) = 1.

(c) By using the fact that the exponential function, x 7→ ex, is continuous show thatlim

n→∞(1 + 1/n)n = e.

4. Show that the function f : [0, π/2] → R defined by, f(x) := sin(x) has a differentiableinverse. Moreover show that (sin−1)′(x) = 1/

√1− x2.

5. Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b). Show that f is aconstant function if, and only if, f ′(x) ≡ 0 on (a, b).

6. Let f : R → R differentiable. Show that if f ′ : R → R is increasing on R then f ′ iscontinuous on R. Hint: use the fact that an increasing function is continuous if, and onlyif, it satisfies the intermediate value property, (see the section on continuity).

7. Let f : R → R and let x0 ∈ R. If f ′′ exists and is continuous on R and (i) f ′(x0) = 0;(ii) f ′′(x0) > 0. Show that f has a local minimum at x0. Hint: Consider the 1st orderTaylor’s expansion of f around x0, (with remainder).

8. (Taylor’s Theorem) Suppose that a < b and f : [a, b] → R. If f (n) is continuous on[a, b] and differentiable on (a, b) then for each x0 ∈ (a, b] there exists a point ζ ∈ (a, x0)such that f(x0) = Pn(x0) + Rn(x0), where

Pn(x0) := f(a) +n∑

k=1

f (k)(a)(x0 − a)k

k!and Rn(x0) := f (n+1)(ζ)

(x0 − ζ)n(x0 − a)

n!

Hint: Consider the auxiliary function g : [a, b] → R defined by,

g(x) := −f(x0) + f(x) +n∑

k=1

f (k)(x)(x0 − x)k

k!+ M · (x0 − x).

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9. Let I := [a, b] and let f : I → R be differentiable on I. Suppose that f(a) < 0 < f(b)and that there exist m, M such that 0 < m 6 f ′(x) 6 M for all x ∈ I. Let x1 ∈ Ibe arbitrary and define xn+1 := xn − f(xn)/M for all n ∈ N. Show that the sequence(xn : n ∈ N) is well defined and converges to the unique zero r ∈ I of f . Hint: ifφ(x) := x− f(x)/M , show that 0 6 φ′(x) 6 1−m/M < 1 and that φ([a, b]) ⊆ [a, b].

10. Calculate the following limits.

(a) limx→0

tan−1(x)

x; (b) lim

x→0

sin(x)

x; (c) lim

x→0

ex − 1

x; (d) lim

x→0

1− cos(x)

x2.

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