Lecture 4 : Calculating Limits using Limit Laws

Lecture 4 : Calculating Limits using Limit Laws

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Using the definition of the limit, limx¡úa f (x), we can derive many general laws of limits, that help us to

calculate limits quickly and easily. The following rules apply to any functions f (x) and g(x) and also

apply to left and right sided limits:

Suppose that c is a constant and the limits

lim f (x)

x¡úa

exist (meaning they are finite numbers).

and

lim g(x)

x¡úa

Then

1. limx¡úa [f (x) + g(x)] = limx¡úa f (x) + limx¡úa g(x) ;

(the limit of a sum is the sum of the limits).

2. limx¡úa [f (x) ? g(x)] = limx¡úa f (x) ? limx¡úa g(x) ;

(the limit of a difference is the difference of the limits).

3. limx¡úa [cf (x)] = c limx¡úa f (x);

(the limit of a constant times a function is the constant times the limit of the function).

4. limx¡úa [f (x)g(x)] = limx¡úa f (x) ¡¤ limx¡úa g(x);

(The limit of a product is the product of the limits).

(x)

x¡úa f (x)

= lim

if limx¡úa g(x) 6= 0;

5. limx¡úa fg(x)

limx¡úa g(x)

(the limit of a quotient is the quotient of the limits provided that the limit of the denominator is

not 0)

Example

If I am given that

lim f (x) = 2,

x¡ú2

lim g(x) = 5,

x¡ú2

lim h(x) = 0.

x¡ú2

find the limits that exist (are a finite number):

2f (x) + h(x) limx¡ú2 (2f (x) + h(x))

=

since lim g(x) 6= 0

x¡ú2

x¡ú2

g(x)

limx¡ú2 g(x)

(a) lim

=

2 limx¡ú2 f (x) + limx¡ú2 h(x)

2(2) + 0

4

=

=

limx¡ú2 g(x)

5

5

f (x)

x¡ú2 h(x)

f (x)h(x)

x¡ú2

g(x)

(b) lim

(c) lim

Note 1 If limx¡úa g(x) = 0 and limx¡úa f (x) = b, where b is a finite number with b 6= 0, Then:

(x)

the values of the quotient fg(x)

can be made arbitrarily large in absolute value as x ¡ú a and thus

1

the limit does not exist.

(x)

If the values of fg(x)

are positive as x ¡ú a in the above situation, then limx¡úa

f (x)

= ¡Þ,

g(x)

(x)

limx¡úa fg(x)

= ?¡Þ,

(x)

If the values of fg(x)

are negative as x ¡ú a in the above situation, then

If on the other hand, if limx¡úa g(x) = 0 = limx¡úa f (x), we cannot make any conclusions about

the limit.

Example Find limx¡ú¦Ð?

cos x

.

x?¦Ð

As x approaches ¦Ð from the left, cos x approaches a finite number ?1.

As x approaches ¦Ð from the left, x ? ¦Ð approaches 0.

Therefore as x approaches ¦Ð from the left, the quotient

cos x

x?¦Ð

approaches ¡Þ in absolute value.

The values of both cos x and x ? ¦Ð are negative as x approaches ¦Ð from the left, therefore

lim?

x¡ú¦Ð

cos x

= ¡Þ.

x?¦Ð

More powerful laws of limits can be derived using the above laws 1-5 and our knowledge of some

basic functions. The following can be proven reasonably easily ( we are still assuming that c is a

constant and limx¡úa f (x) exists );



n

6. limx¡úa [f (x)]n = limx¡úa f (x) , where n is a positive integer (we see this using rule 4 repeatedly).

7. limx¡úa c = c, where c is a constant ( easy to prove from definition of limit and easy to see from

the graph, y = c).

8. limx¡úa x = a, (follows easily from the definition of limit)

9. limx¡úa xn = an where n is a positive integer (this follows from rules 6 and 8).

¡Ì

¡Ì

10. limx¡úa n x = n a, where n is a positive integer and a > 0 if n is even. (proof needs a little extra

work and the binomial theorem)

p

p

11. limx¡úa n f (x) = n limx¡úa f (x) assuming that the limx¡úa f (x) > 0 if n is even. (We will look at

this in more detail when we get to continuity)

Example Evaluate the following limits and justify each step:

(a)

limx¡ú3

(b)

limx¡ú1

x3 +2x2 ?x+1

x?1

¡Ì

3

x+1

2

(c)

Determine the infinite limit (see note 1 above, say if the limit is ¡Þ, ?¡Þ or D.N.E.)

x+1

limx¡ú2? (x?2)

.

Polynomial and Rational Functions

Please review the relevant parts of Lectures 3, 4 and 7 from the Algebra/Precalculus review

page. This demonstration will help you visualize some rational functions:

Direct Substitution (Evaluation) Property If f is a polynomial or a rational function and

a is in the domain of f , then limx¡úa f (x) = f (a). This follows easily from the rules shown above.

(Note that this is the case in part (a) of the example above)

if f (x) =

P (x)

Q(x)

is a rational function where P (x) and Q(x) are polynomials with Q(a) = 0, then:

P (x)

If P (a) 6= 0, we see from note 1 above that limx¡úa Q(x)

= ¡À¡Þ or D.N.E. and is not equal to ¡À¡Þ.

If P (a) = 0 we can cancel a factor of the polynomial P (x) with a factor of the polynomial Q(x)

and the resulting rational function may have a finite limit or an infinite limit or no limit at x = a.

P (x)

by the following observation which

The limit of the new quotient as x ¡ú a is equal to limx¡úa Q(x)

we made in the last lecture:

Note 2: If h(x) = g(x) when x 6= a, then limx¡úa h(x) = limx¡úa g(x) provided the limits exist.

Example

to ¡À¡Þ:

Determine if the following limits are finite, equal to ¡À¡Þ or D.N.E. and are not equal

(a) limx¡ú3

x2 ?9

.

x?3

(b)

limx¡ú1?

x2 ?x?6

.

x?1

(c) Which of the following is true:

2 ?x?6

1. limx¡ú1 x x?1

= +¡Þ,

2. limx¡ú1

¡À¡Þ,

x2 ?x?6

x?1

3

= ?¡Þ,

3. limx¡ú1

x2 ?x?6

x?1

D.N.E. and is not

Example

Evaluate the limit (finish the calculation)

(3 + h)2 ? (3)2

.

h¡ú0

h

lim

limh¡ú0

(3+h)2 ?(3)2

h

Example

= limh¡ú0

9+6h+h2 ?9

h

=

Evaluate the following limit:

¡Ì

lim

x¡ú0

x2 + 25 ? 5

.

x2

Recall also our observation from the last day which can be proven rigorously from the definition

(this is good to keep in mind when dealing with piecewise defined functions):

Theorm limx¡úa f (x) = L

if and only if

limx¡úa? f (x) = L = limx¡úa+ f (x).

Example Evaluate the limit if it exists:

3x + 6

x¡ú?2 |x + 2|

lim

The following theorems help us calculate some important limits by comparing the behavior of a

function with that of other functions for which we can calculate limits:

4

Theorem If f (x) ¡Ü g(x) when x is near a(except possible at a) and the limits of f (x) and g(x)

both exist as x approaches a, then

lim f (x) ¡Ü lim g(x).

x¡úa

x¡úa

The Sandwich (squeeze) Theorem

If f (x) ¡Ü g(x) ¡Ü h(x) when x is near a (except

possibly at a) and

lim f (x) = lim h(x) = L

x¡úa

x¡úa

then

lim g(x) = L.

x¡úa

Recall last day, we saw that limx¡ú0 sin(1/x) does not exist because of how the function oscillates near x = 0. However we can see from the graph below and the above theorem that

limx¡ú0 x2 sin(1/x) = 0, since the graph of the function is sandwiched between y = ?x2 and

1

y = x2 :

0.5

K1

0

K0.5

0.5

x

1

K0.5

Example Calculate the limit limx¡ú0 x2 sin x1 . K1

We have ?1 ¡Ü sin(1/x) ¡Ü 1O for all x,

multiplying across by x2 (which is positive), we get ?x2 ¡Ü x2 sin(1/x) ¡Ü x2 for all x,

Using the Sandwich theorem, we get

0 = lim ?x2 ¡Ü lim x2 sin(1/x) ¡Ü lim x2 = 0

x¡ú0

x¡úo

x¡ú0

Hence we can conclude that

lim x2 sin(1/x) = 0.

x¡ú0

Example

Decide if the following limit exists and if so find its values:

limx¡úo x100 cos2 (¦Ð/x)

5

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