calculus applied to business part03

Limits and Continuity

Calculus Applied to Business and EconomicsLimits and Continuity

Prof. Kenneth James T. Nuguid

April 29, 2013

Prof. Kenneth James T. Nuguid Calculus Applied to Business and Economics


Limits

One important aspect of the study of calculus is the analysis ofhow function values (outputs) change as input values change.Basic to this study is the notion of a limit.

Limit of a Function

Suppose a function f is given and suppose the x-values (theinputs) get closer and closer to some number a. If thecorresponding outputsthe values of f (x)get closer and closer toanother number L, then that number L is called the limit of as xapproaches a. This is denoted by

limxa

f (x) = L.

Example 1. Find the limit as x approaches 4 for f (x) = 2x + 3.



Limits



Limits

The symbol stands for approaches from either side. Thus weshall adopt the following notation when necessary:

limxa

f (x) to indicate limit from the left (x < a)

orlim

xa+f (x) to indicate limit from the right (x > a).

Remarks

i. These are called left-hand and right-hand limits.

ii. In order for a limit to exist, both the left-hand and right-handlimits must exist and be the same.

iii. The limit L must be a unique real number.



Limits

Example 2. Let f (x) = x21x1 .

a. What is f (1)?

b. What is the limit of f (x) as x approaches 1?

Solution:

a. There is no answer, since we get a 0 in the denominator:

f (1) =12 11 1

=0

0.

Thus, f (1) does not exist or is undefined. note that there is nopoint on the graph of f (x) shown on the next slide for x = 1

b. We select x-values close to 1 on either side and and we see thatthe function values get closer and closer to 2. Thus,

limx1

f (x) = 2.



Limits

Figure : The graph has a hole at the point (1, 2), Thus, even though thefunction is not defined at x = 1, the limit does exist as x 1.



Limits

Example 3. Consider a piecewise function H given by

H(x) =

{2x + 2 if x < 1,2x 4 if x 1.

a. Find each of the following limits, if they exist.

i. limx1 H(x)ii. limx1+ H(x)

b. Graph the function.

When necessary, state that the limit does not exist.



Limits

a. We examine the values of H(x) through the table below:

Figure : Table of values for the LHL and RHL.

Based from the table above,

i. limx1 H(x) = 4

ii. limx1+ H(x) = 2Prof. Kenneth James T. Nuguid Calculus Applied to Business and Economics


Limits

Since the limit from the left, 4, is not the same as the limit fromthe right, 2, we say that limx1 H(x)does not exist.

b. The graph of H(x) is shown below:

We note that H(1) = 2. Inthis example, the functionvalue exists for x = 1, but thelimit as x approaches 1 doesnot exist.



Limits

Remark

The limit at a number does not depend on the function value at aor even on whether that function value, f (a) exists. That is,whether or not a limit exists at a has nothing to do with thefunction value f (a).

Exercise 1. Find the limit of f (x) = 3x 1 as x approaches 6.Exercise 2. Find the limit of f (x) = x

29x3 as x approaches 3. Is

this limit equal to f (3)?Exercise 3. Let

k(x) =

{x + 4 if x 3,2x + 1 if x > 3.

Find the LHL and RHL as x approaches 3. Does the limit as xapproaches 3 exist?



Limits Involving Infinity

Example 4a. Let f (x) = 1x .

a. What is f (0)?

b. Find limx0 f (x).c. Find limx0+ f (x).

Solution:

a. f (0) does not exist; there is no point on the graph thatcorresponds to x = 0.

b. We can use the table below tofind the LHL.

Figure : LHL =

c. We can use the table below tofind the RHL.

Figure : RHL =




Since the left-hand and right-hand limits do not match (and arenot finite), limx0 f (x) does not exist.Now, let us see what happens when the inputs x are the onesincreasing or ddecreasing without bounds. Such limits are calledlimits at infinity and are written as follows:

limx

f (x) and limx

f (x).

Example 4b. Compute the limits at infinity for f (x) = 1x . Thengraph f (x). Solution:




Exercise 4. Let

f (x) =1

x 2+ 3.

Find the limit as x approaches 2 (LHL and RHL) and the limits atinfinity then graph.



Limits Properties

The following list summarizes common limit properties that allowus to calculate limits much more efficiently.

Limit Properties

If limxa f (x) = L and limxa g(x) = M and c is a constant, thenwe have the following:

i. The limit of a constant is the constant:

limxa

c = c .

ii. The limit of a power is the power of that limit.

limxa

[f (x)]n = [ limxa

f (x)]n = Ln.



Limits Properties

Limit Properties

iii. The limit of a root is the root of that limit (assuming n is apositive integer), where L 0 if n is even:

limxa

n

f (x) = n

limxa

f (x) =n

L.

iv. The limit of a sum or a difference is the sum or the differenceof the limits:

limxa

[f (x) g(x)] = limxa

f (x) limxa

g(x) = LM.

v. The limit of a product is the product of the limits:

limxa

[f (x) g(x)] = limxa

f (x) limxa

g(x) = L M.



Limits Properties

Limit Properties

vi. The limit of a quotient is the quotient of the limits, whereM 6= 0:

limxa

f (x)

g(x)=

limxa f (x)

limxa g(x)=

L

M.

vii. The limit of a constant times a function is the constant timesthe limit of the function:

limxa

[c f (x)] = c limxa

f (x) = c L.



Limits Properties

Example 5a. Use the Limit Properties to find

limx4

(x2 3x + 7).

Solution: limxa x = a implies that limx4 x = 4.So by property (v),

limx4

x2 = limx4

x limx4

x = 4 4 = 16.

By limit property 7,

limx4

(3x) = 3 limx4

x = 3 4 = 12.

By limit property 1limx4

7 = 7.

Combining the above results and using limit property 4,

limx4

(x2 3x + 7) = 16 12 + 7 = 11.



Limit Properties

Theorem on Limits of Rational Functions

For any rational function F, with a in the domain of F,

limxa

F (x) = F (a).

Rational functions are a family of common functions, including allpolynomial functions (which include constant functions and linearfunctions) and ratios composed of such functions.

Example 5b. Find limx2 (x4 5x3 + x2 7).

Solution: It follows from the Theorem on Limits of RationalFunctions that we can find the limit by substitution:

limx2

(x4 5x3 + x2 7) = 24523+227 = 1640+47 = 27.



Limit Properties

Example 5c. Find limx0

x2 3x + 2.Solution: It follows from the Theorem on Limits of RationalFunctions and limit property 3 that:

limx0

x2 3x + 2 =

limx0

(x2 3x + 2) =

02 3 0 + 2 =

2.

Exercise. Find the following limits.5a. limx1 2x

3 + 3x2 65b. limx4

2x2+5x13x2

5c. limx2

1 + 3x2



Limit Properties

Example 6. Let r(x) = x2x12x+3 . Find

limx3

x2 x 12x + 3

.

Solution: We note that r(3) does not exist. We can verify bytabular and graphical method that the limit exists and is equal to7. This can be supported algebraically by

limx3

x2 x 12x + 3

= limx3

(x + 3)(x 4)(x + 3)

= limx3

x 4 = 34 = 7.

Exercise 6. Evaluate:

limx2

x2 + 4x 12x2 4



Limit Properties

Example 7. Evaluate:

limh0

3x2 + 3xh + h2.

Solution: We treat x as a constant since we are interested only inthe way in which the expression varies when h approaches 0.

We use the Limit Properties to find that

limh0

3x2 + 3xh + h2 = 3x2 + 3x(0) = 02 = 3x2.

Exercise 7. Evaluate:

limh0

2x h(x + h)2x2



Continuity

Functions that can be traced on the coordinate system withoutlifting your pencil or pen are said to be continuous functions.The functions below are continuous at any real number.

Figure : Note that there are no jumps or holes in the graphs.



Continuity

In each case, the graph cannot be traced without lifting the pencilfrom the paper and hence, are branded as discontinuousfunctions. The point at which the pencil is lifted is a point ofdiscontinuity. See the graphs below:

Figure : Each of these funcitons has a point of discontinuity at whichthere is a jump, break or hole.



Continuity

Continuity at a Point

A function f (x) is continuous at x = a if:

i. f (a) exists. (The output at a exists.)

ii. limxa f (x) exists. (The limit as x a exists.)iii. f (a) = limxa f (x). (The limit is the same as the output.)

A function is continuous over an interval I if it is continuous ateach point in I.



Continuity

Example 8. Determine whether the function f (x) = 2x + 3 iscontinuous at x = 4.Solution: This function is continuous at x = 4 because:

i. f (4) exists. (f (4) = 11.)ii. limx4 f (x) exists. (This limit was found earlier to be 11.)iii. f (4) = limx4 f (x) = 11. (The limit is the same as the

output.)

Exercise 8. Determine the continuity of the following functions atthe given point.

i. f (x) = x2 5 at x = 3.ii.

g(x) =

{12x + 3 if x < 2,

x 1 if x 2, at x = 2.

iii.

F (x) =

{x216x4 if x 6= 4,

7 if x = 4,at x = 4.



Continuity

Application: Price Breaks

Example 9a. Ricks Rocks sells decorative landscape rocks in bulkquantities. For quantities up to and including 500 lb, Rick chargesP25 per pound. For quantities above 500 lb, he charges P20 perpound. The price function can be stated as a piecewise function:

p(x) =

{25x if 0 x 500,20x if x > 500,

where p(x) is the price in pesos for x lbs of rocks. Is the pricefunction continuous at x = 500? Why or why not?



Continuity

Solution:

Figure : Graph of p(x) inexample 9a

The graph shows that

limx500

p(x) = 12, 500

but

limx500+

p(x) = 10, 000.

Since the LHL and the RHLare not equal, then the limit ofp(x) as x approaches 500 doesnot exist.Therefore, there is a pricebreak.



Continuity

Application: Price Breaks

Example 9b. Rick, of Ricks Rocks in Example 9a, realizes that hiscustomers are taking advantage of him: for example, they pay lessfor 550 lb of rocks than they would for 500 lb of rocks. For Rick,this means lost revenue, so he decides to add a quantity discountsurcharge for quantities above 500 lb. If k represents thissurcharge, the price function becomes:

p(x) =

{25x if 0 x 500,

20x + k if x > 500.

Find k such that the function is continuous at x = 500.



Continuity

Solution: For p(x) to be continuous at x = 500,

limx500

p(x) = limx500+

p(x)

= limx500+

20x + k

= 20(500) + k = 12, 500.

Hence, k = 12, 500 10, 000 = 2, 500.



Continuity

Exercise 9. The Candy Factory sells candy by the pound, chargingP15 per pound for quantities up to and including 20 pounds.Above 20 pounds, the Candy Factory charges P12.50 per poundfor the entire quantity, plus a quantity surcharge k. If x representsthe number of pounds, the price function is

p(x) =

{15x if 0 x 20,

12.50x + k if x > 20.

a. Why is there a price break at x = 20?

b. Find k such that the price function p is continuous at x = 20?

c. Why is it preferable to have continuity at x = 20.


calculus applied to business part03

Documents

limit of f x

right x

graph of f x

limx1f x

left x aorlimxa f x

xvalues theinputs

kenneth james

nuguid calculus