Notes, part Notes, part 55

L’HospitalL’HospitalAnother useful technique for computing limits

is L'Hospital's rule:

Basic version:

0)(lim)(lim

xgxfaxax

)(lim)(lim xgxfaxax

)('

)('lim

)(

)(lim

xg

xf

xg

xfaxax

If , then

provided the latter exists.

This also applies if

Fancy L’HospitalFancy L’Hospital

s.often work rule sHospital'L' applying

then and logarithm thegthen takin

,1or or 0 like lookslimit theIf 00

You can use “Basic L’Hospital” for #4, 6, 8 and 10 of the top ten limit list. But for limits like #9, you need...

Top ten famous limits:Top ten famous limits:

1

lim0 xx

1

lim0 xx

0 1

lim xx

1.

2.

3. (A) If 0 < x < 1 then 0lim

n

nx

(B) If x > 1, then

n

nxlim

4. and1sin

lim0

x

xx

0cos1

lim0

x

xx

0 e lim

x

x5. and

e lim x

x

6-106-10

6. For any value of n,

and for any positive value of n,

7.

1sin lim

0

xx

does not exist!

0lim x

n

x e

x

0ln

lim nx x

x

0)ln( lim0

xxx

ex

xx

11 lim

8.

9.

)(')()(

lim afax

afxfax

10. If f is differentiable at a, then

20

cos1lim

x

xx

564

32lim

2

2

xx

xxx

xxxx

1lim 2

Here are three more:Here are three more:

A challenge:

How about this one?How about this one?

2/1

0x

sinlim

x

x

x

A.

B. 0

C. -

D. 1

E.

F.

G.

H.

2e

2

6/1e

4/1e

Last one Last one (for now)...(for now)...

A.

B. 1/2

C.

D. 3

E. F. undefined

G.

H. 22e

2e2e

22 e

23e

xtx

xdtexe

0

22

lim

Improper Improper integralsintegrals

These are a special kind of limit. An improper integral is one where either the interval of integration is infinite, or else it includes a singularity of the function being integrated.

Examples of the Examples of the first kind are:first kind are:

dxx

dxe x

0 -21

1 and

Examples of the second kind:Examples of the second kind:

32

4

tan and 11

0

dxxdxx

The second of these is subtle because the singularity of tan x occurs in the interior of the interval of integration rather than at one of the endpoints.

Same methodSame methodNo matter which kind of improper integral (or combination of improper integrals) we are faced with, the method of dealing with them is the same:

b

x

b

x dxedxe00

lim as thingsame themeans

Calculate the limit!Calculate the limit!What is the value of this limit

(and hence, of the improper integral )?

A. 0

B. 1

C.

D.

E.

ebe

dxe x

0

Another improper integralAnother improper integral

?1

1 of value theisWhat

.arctan1

1 that Recall

-2

2

dxx

Cxdxx

A. 0

B.

C.

4

2

D.

E.

Area between the x-Area between the x-axis and the graphaxis and the graph

The integral you just worked represents all of the area

between the x-axis and the graph of 211x

The other type...The other type...For improper integrals of the other type, we make the same kind of limit definition:

.1

lim be todefined is 1 1

0

1

0

dxx

dxx a

a

Another example:Another example:What is the value of this limit, in

other words, what is

A. 0

B. 1

C. 2

D.

E.

2

?11

0

dxx

A divergent improper integralA divergent improper integral

It is possible that the limit used to define the improper integral fails to exist or is infinite. In this case, the improper integral is said to diverge . If the limit does exist, then the improper integral converges. For example:

1

00

ln1ln lim1

adxx a

so this improper integral diverges.

One for you:One for you:

3

2

4

diverge?or converge )tan( Does

dxx

A. Converge

B. Diverge

Sometimes it is possible...Sometimes it is possible...

to show that an improper integral converges without actually evaluating it:

So the limit of the first integral must be finite as b goes to infinity, because it increases as b does but is bounded above (by 1/3).

.1 allfor 3

11

7

1

thathave we,0 allfor 1

7

1 Since

331

1 144

44

bb

dxx

dxxx

xxxx

b b

A puzzling example...A puzzling example...Consider the surface obtained by rotating the graph

of y = 1/x for x > 1 around the x-axis:

Let’s calculate the volume contained inside the surface:

units. cubic 1 lim 1

1

21

bbx dxV

What about the surface area?What about the surface area?This is equal to...

1

41

2 11

2)('1)(2 dx

xxdxxfxfSA

This last integral is difficult (impossible) to evaluate directly, but it is easy to see that its integrand is bigger than that of the divergent integral

Therefore it, too is divergent, so the surface has infinite surface area.

This surface is sometimes called "Gabriel's horn" -- it is a surface that can be "filled with water" but not "painted".

dxx

1

2

SequencesSequences

The lists of numbers you generate using a numerical method like Newton's method to get better and better approximations to the root of an equation are examples of (mathematical) sequences .

Sequences are infinite lists of numbers, Sometimes it is useful to think of them as functions from the positive integers into the reals, in other words,

,...,, 321 aaa

forth. so and ,a(2) ,a(1) 21 aa

The feeling we have about numerical methods like the bisection method, is that if we kept doing it more and more times, we would get numbers that are closer and closer to the actual root of the equation. In other words

where r is the root.

Sequences for which exists

and is finite are called convergent, other sequences are called divergent

Convergent and DivergentConvergent and Divergent

nn

a lim

rann

lim

For example...For example...

The sequence

1, 1/2, 1/4, 1/8, 1/16, .... , 1/2 , ... is convergent (and converges to zero, since

), whereas:

the sequence 1, 4, 9, 16, .…n , ... is divergent.

2

n

0 lim21

n

n

PracticePractice

The sequence

...,2

1,......,

5

4,

4

3,

3

2

n

n

A. Converges to 0

B. Converges to 1

C. Converges to n

D. Converges to ln 2

E. Diverges

Another...Another...

The sequence

...,2

1)1(,......,

5

4,

4

3,

3

2

n

nn

A. Converges to 0

B. Converges to 1

C. Converges to -1

D. Converges to ln 2

E. Diverges

A powerful existence theoremA powerful existence theoremIt is sometimes possible to assert that a sequence is convergent even if we can't find the limit right away. We do this by using the least upper bound property of the real numbers:

If a sequence has the property that a <a <a < .... is called a "monotonically increasing" sequence. For such a sequence, either the sequence is bounded (all the terms are less than some fixed number) or else it increases without bound to infinity. The latter case is divergent, and the former must converge to the least upper bound of the set of numbers {a , a , ... } . So if we find some upper bound, we are guaranteed convergence, even if we can't find the least upper bound.

1 2 3

1 2

Consider the sequence...Consider the sequence...

222,22 ,2

To get each term from the previous one, you add 2 and then take the square root.

It is clear that this is a monotonically increasing sequence. It is convergent because all the terms are less than 2. To see this, note that if x>2, then

So our terms can't be greater than 2, since adding 2 and taking the square root makes our terms bigger, not smaller.

Therefore, the sequence has a limit, by the theorem.

etc.

.2 so and ,222 xxxxx

QUESTION:QUESTION:

What is the limit?

...,222,22 ,2

Newton’s MethodNewton’s MethodA better way of generating a sequence of numbers that are (usually) better and better solutions of an equation is called Newton's method.

In it, you improve a guess at the root by calculating the place where the tangent line drawn to the graph of f(x) at the guess intersects the x-axis. Since the tangent line to the graph of y = f(x) at x = a is y = f(a) + f '(a) (x-a), and this line hits the x-axis when y=0, we solve for x in the equation f(a) + f '(a)(x-a) = 0 and get x = a - f(a)/f '(a).

)('

)(

old

oldoldnew xf

xfxx

Let’s try itLet’s try it

on the same function we used before,

with the guess that the root x1 = 2. Then the next guess is

This is 1.8. Let's try it again. A calculator helps:

22)( 3 xxxf

5

9

10

22

)('

)(

1

112

xf

xfxx

76995.1)8.1('

)8.1(8.1

)('

)(

2

223

f

f

xf

xfxx

We’re already quite close...We’re already quite close...

with much less work than in the bisection method! One more time:

And according to Maple, the root is

fsolve(f(x)=0);

So with not much work we have the answer to six significant figures!

1.769292354

769292663.1)76005.1('

)76995.1(76995.1

)('

)(

3

334

f

f

xf

xfxx

Your turn…Your turn…

Try Newton's method out on the equation

First make a reasonable guess, then iterate. Report your answer when you get two successive iterations to agree to five decimal places.

0325 xx

FractalsFractals

Fractals are geometric figures constructed as a limit of a sequence of geometric figures.

Koch Snowflake

Sierpinski Gasket

Newton's method fractals