example of computing limits using Taylor expansion

example of computing limits using Taylor expansion

Problem

.

Evaluate

lim

x

→

-

∞

⁡

(

x

+

x

2

+

23

⁢

x

)

.

In beginners’ courses in

calculus

, one is usually told to
rationalize the above

expression

by multiplying by

(

x

-

x

2

+

23

⁢

x

)

/

(

x

-

x

2

+

23

⁢

x

)

.
This approach is somewhat dissatisfying, because it depends
on a specific

algebraic

trick that works only for square roots
— for example, it would have been more difficult or impossible to rationalize, if instead, we had a cube root or even a

transcendental function

—
and this trick does not appeal to our intuition that

x

2

+

23

⁢

x

should be approximately

|

x

|

for large

|

x

|

.
Fortunately, there is another approach that
exploits the

analytic

properties

of the

functions

involved.

L’Hôpital’s Rule (

http://planetmath.org/LHpitalsRule

) is one analytic approach,
but in many cases, using

Taylor expansion

is even easier
and straightforward. Essentially, Taylor expansion approximates
complicated functions by

polynomials

, whose limits are easy to evaluate.
We illustrate the method below.

First rewrite

lim

x

→

-

∞

⁡

(

x

+

x

2

+

23

⁢

x

)

=

lim

x

→

+

∞

⁡

(

x

2

-

23

⁢

x

-

x

)

,

so that we do not have to worry about the pesky

negatives

any more.
Then, with the help of the

binomial formula

:

(

1

+

y

)

1

/

2

=

1

+

1

2

⁢

y

+

o

⁢

(

y

)

,

as

y

→

0

,

(“

o

” is

Landau notation

)
we obtain:

x

2

-

23

⁢

x

-

x

=

x

⁢

(

1

-

23

x

-

1

)

=

x

⁢

(

1

-

1

2

⁢

23

x

+

o

⁢

(

23

x

)

-

1

)

,

as

x

→

∞

(so

y

=

-

23

x

→

0

)

=

-

23

2

+

x

⁢

o

⁢

(

23

x

)

=

-

23

2

+

o

⁢

(

1

)

.

Therefore

lim

x

→

-

∞

⁡

(

x

+

x

2

+

23

⁢

x

)

=

lim

x

→

∞

⁡

(

x

2

-

23

⁢

x

-

x

)

=

-

23

2

.

Problem

.

Evaluate

lim

x

→

0

⁡

(

1

ln

⁡

(

1

+

x

)

-

1

tan

⁡

x

)

.

This example is admittedly artificial; it was made to be annoying to solve using
L’Hôpital’s Rule alone, but much simpler if one knows how to use
the Taylor expansions:

ln

⁡

(

1

+

x

)

=

x

-

1

2

⁢

x

2

+

o

⁢

(

x

2

)

,

as

x

→

0

, and

tan

⁡

x

=

x

+

o

⁢

(

x

2

)

,

as

x

→

0

.

So we compute:

1

ln

⁡

(

1

+

x

)

-

1

tan

⁡

x

=

tan

⁡

x

-

ln

⁡

(

1

+

x

)

ln

⁡

(

1

+

x

)

⁢

tan

⁡

x

=

(

x

+

o

⁢

(

x

2

)

)

-

(

x

-

1

2

⁢

x

2

+

o

⁢

(

x

2

)

)

ln

⁡

(

1

+

x

)

⁢

tan

⁡

x

=

1

2

⁢

x

2

+

o

⁢

(

x

2

)

(

x

+

o

⁢

(

x

)

)

⁢

(

x

+

o

⁢

(

x

)

)

=

1

2

+

o

⁢

(

1

)

(

1

+

o

⁢

(

1

)

)

⁢

(

1

+

o

⁢

(

1

)

)

Therefore,

lim

x

→

0

⁡

(

1

ln

⁡

(

1

+

x

)

-

1

tan

⁡

x

)

=

1

2

.

The reader might reasonably ask how
did we know the right number of terms to use in the Taylor expansions.
The answer is to guess.
This is not as problematic as it sounds. First, there is no harm
in using more terms than necessary in the expansion (only that there is more writing).
And if we used too few terms, we would know when we later encounter indeterminate forms such as

o

⁢

(

1

)

/

x

(as

x

→

0

) in our

derivations

. If that happens,
it is not hard to go back and add the needed terms.

Notice that all the essential information to evaluate the limit is contained in the first few

derivatives

of
the functions involved

at particular points

— in the above example,
only at

x

=

0

.
This information can be obtained by manipulating series,
unlike L’Hôpital’s Rule
which necessitates computing the derivative

functions

at all points.
So even monstrous expressions like this one is tractable
with Taylor expansion:

lim

x

→

0

⁡

(

1

ln

⁡

(

1

+

tan

⁡

(

sin

⁡

x

)

)

-

1

e

sin

⁡

(

tan

⁡

x

)

-

1

)

.

On the other hand, there definitely are situations where
L’Hôpital’s Rule works but Taylor expansion does not: for instance,

lim

x

→

0

+

⁡

x

⁢

ln

⁡

x

=

lim

x

→

0

+

⁡

x

1

/

ln

⁡

x

,

because

1

/

ln

⁡

x

cannot be expanded in a

Taylor series

about

x

=

0

.

Problem

.

Here is a problem involving a different sort of limit:
does the following series

converge

?

∑

n

=

1

∞

arctan

⁡

(

n

-

1

)

Our intuition suggests that it does not, because

arctan

⁡

(

n

-

1

)

should be approximately

n

-

1

, and

∑

n

n

-

1

diverges.
However, the standard

comparison test

does not work
because

arctan

⁡

x

≤

x

(for

x

≥

0

) has the

inequality

in the wrong direction.
But with Taylor expansion the

solution

is a snap.
By expanding

arctan

⁡

x

=

x

+

O

⁢

(

x

3

)

,

as

x

→

0

.

and

summing

both sides, we get

∑

n

=

1

∞

arctan

⁡

(

n

-

1

)

=

∑

n

=

1

∞

n

-

1

+

∑

n

=

1

∞

O

⁢

(

n

-

3

)

.

As

∑

n

O

⁢

(

n

-

3

)

converges (being dominated by

C

⁢

∑

n

n

-

3

for some constant

C

),

∑

n

arctan

⁡

(

n

-

1

)

must diverge (to

∞

).

(Of course, this problem could be solved by using the

integral test

,
but who really wants to

integrate

∫

arctan

⁡

(

1

/

x

)

⁢

𝑑

x

?)

Title

example of computing limits using Taylor expansion

Canonical

name

ExampleOfComputingLimitsUsingTaylorExpansion

Date of creation

2013-03-22 15:39:48

Last modified on

2013-03-22 15:39:48

Owner

stevecheng (10074)

Last modified by

stevecheng (10074)

Numerical id

15

Author

stevecheng (10074)

Entry type

Example

Classification

msc 26A06

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