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1 - limits notes
Industrial Engineering (ERGO1)
University of the Assumption
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After defining and graphing functions, we are now in position to tackle one of the origins of the concept of calculus. We develop the principle by exploring some other concepts of functions.
Suppose we are given with the function
####### ø ù
1
232
x
xf xx.
####### Note that the function given is not defined when x 1 since the value of ø ù xf will be undefined.
Automatically, the value of 1 is discarded from its domain.
But even though we cannot allow 1 as our value of x , what if we set the values of x to be
####### closer and closer to 1? Let us try to see on what it will seem to be the value of ø ù xf if we set the
value of x to be very close to 1, but not necessarily equal to 1.
To answer this question, let us construct a table of values with x- values that are less than 1 (0, 0, 0, 0) and greater than 1 (1, 1, 1, 1). Then we substitute these
####### values to ø ù
1
232
x
xf xx using a calculator. Setting up the table and solving for the
####### corresponding values of ø ù xf , we have:
x 0 0 0 0 1 1 1 1 1.
####### ø ù xf 4 4 4 4 und 5 5 5 5.
Let us deduce all possible concepts that we can notice in the table of values. Starting with values of x that are less than 1 (0, 0, 0, 0), note that as the values come closer and
####### closer to 1, the values of ø ù xf seem to come closer and closer to a certain value which is 5.
Notice that the outcome is the same for values of x that are greater than 1 (1, 1, 1,
####### 1), that is as the values of x comes nearer and nearer to 1, the value of ø ù xf also seem to
####### come closer and closer to 5. But again, note that the value of ø ù xf is NOT equal to 5 if x is equal
to 1 because in the first place, the function cannot take a value of x = 1. The phrases <comes closer and closer= and <equal to= are two different things and have different meanings.
To support these results, let us sketch the graph of the involved function using the concepts that we learned in graphing graphs of functions. Noting that the numerator is factorable :
1. THE LIMIT OF A FUNCTION
####### ø ù
1
232
x
xf xx ; x 1
####### ø ùø ù 32
1
321
x x
xx
####### ø ù xxf ü 32 ; x 1 (graph is a straight
####### line , replace ø ù xf by y )
xy 32 ; m = 2 and b = 3 (0, 3)
####### *if x 1 ; y ø ù 32312
y 5
restricted point is at (1, 5)
Note that our mathematical manipulation and graph matches with our findings earlier. That
####### is, as the value of x comes closer and closer to 1, the value of ø ù xf comes closer and closer to
5. But again the value of x cannot be equal to 1 and the value of y cannot be equal to 5.
In mathematical terms, the phrase <closer and closer= is synonymous to the term < approaches =. To rephrase our findings, we have this statement: <As the value of x approaches
####### 1, the value of ø ù xf approaches 5=. Symbolically, we can symbolize the term <approaches= in
terms of an arrow pointing to the right (). Then x approaches 1 can be symbolized as x 1.
####### The number 5 wherein it is the value that ø ù xf approaches is called the limit of the function
####### ø ù xf , with this formal definition:
It is very important to take note that in finding the limit ( L ) of the function, we just consider
####### on where the value of ø ù xf seems to be close as the value of x comes closer and closer to
the value of a , but not necessarily equal to a. Oftentimes, the value of a is the restricted value of x , so the value of x cannot be equal to a , just like in our illustration, that is x cannot be equal to 1.
Definition : Limit of a Function
Let be a function defined at every number in some interval containing a , except possibly at the number a itself. The limit of a as x approaches a is L , written as:
a. x 6 ü 77lim b. x 8476 ü 2121lim
Limit Theorem # 2: The Limit of an Identity Function
####### If a is a constant, then if ø ù xxf ,
lim ax ax
####### This limit is named as the limit of an <identity function= since if ø ù xf is replaced by y , the
expression involved will be xy where the values of x and y are just equal or identical.
According to this limit theorem , if the function involved is just a variable raised to 1 and having a numerical coefficient of 1, the limit is just equal to a , or the value where the variable involved is approaching into.
Example 3 : Evaluate x lim 4 x
Note that the function involved is just x, so it is just a limit of an identity function. Whenever we have this case that the function involved is just a variable having an exponent of 1 and numerical coefficient of 1, the limit is always equal to a in general, or:
x 4 x ü 4lim
Example 4 : Evaluate (a) z lim 9 z ; (b) q lim 576 q
The variables involved is not x , but since the variables involved in each of the function involved and the variable under the <lim= notation are just the same, we can apply limit theorem 2:
a. z 9 z ü 9lim b. q 576 q ü 576lim
The next limit theorem will illustrate the process of finding the limit of functions having more than one term , that is if each term is separated by a plus or minus sign.
Limit Theorem # 3: The Limit of a Sum and/or Difference of n Functions
####### If lim ax ø ù Lxf 11 ; lim ax ø ù Lxf 22 ; lim ax ø ù Lxf 33 ; ... ; lim ax ø ù Lxf nn , then:
ax û ø ù ø ù ø ù ø ùý 32 ..... LLLLxfxfxfxf n1n
The statement of this limit theorem seems to very complicated because of numerous symbols and expressions. But we can summarize the statement of this limit theorem as follows: if we are involved with the limit of a sum or difference of 2 or more terms, we can just distribute the limit expression to the terms and evaluate the individual limits. That is, in general:
ax û ø ù ø ù ø ù n321 ø ùý ax 1 ø ù ax 2 ø ù ax 3 ø ù lim...limlimlim.. ax n ø ù xf xf xf xf xfxfxfxf
32 ... LLLL n
####### Example 5 : Evaluate x 3 ø ù x 6lim
The expression that we will find the limit is composed of 2 terms. As a first step, we can apply limit theorem # 3 by distributing the limit expression to the two terms. Doing this:
####### ø ù xxx limlim6lim 333 6 x x
Applying next limit theorem # 2 in the first term and limit theorem # 1 in the second term:
####### ø ù xxx 333 6 x x 63limlim6lim
####### x 3 ø ù x ü 36lim
####### Example 6 : Evaluate y 7 ø y 19lim ù
Note that this example is somewhat similar in form as the previous example. Executing the steps and applying the limit theorems that we know so far::
####### ø ù yyy 777 y y 19719limlim19lim
####### y 7 ø y ùü 1219lim
WARNING : In evaluating limits, unnecessary shortcut solutions are strictly not allowed. If we are involved with evaluating limits of more than one term, we are required to distribute first the limit expression to each term before we use the concept of the limit of a constant and limit of an identity function. The use of too much shortcuts can lead to violation of concepts in
x 2 ø x ù ø ùø ù 7672373lim
x 2 ø x ù ü 1373lim
WARNING : It is strongly emphasized that in evaluating limits of functions, unnecessary shortcuts are not allowed. In our previous example, it seems that the answer of -13 can also be obtained by substituting -2 to the value of x and simplifying. Recall that the concept of evaluating the limit of any function is we let the value of x to be closer and closer to a certain number (in this case, closer and closer to -2) BUT NOT NECESSARILY EQUAL TO THAT NUMBER. Substituting -2 to x means that we let the value of x to be equal to -2, which is incorrect considering that we are evaluating the limit. To illustrate this error in evaluating limits:
x 2 ø x ù ø ù 137672373lim
What we have done there is we did not evaluate the limit. The solution that we have done is
for finding an expression for f ø ù 2 by substitution. That is in general, the answer for f ø ù 2 is
not always equal to the expression for x lim 2 ø ù xf. It is a mortal sin to substitute the value that
x is approaching in evaluating limits. Inventing our own concepts in finding limits is not allowed, the only process for this is by using limit theorems.
Example 8 : Evaluate x 4 ø ùø xx 9234lim ù
The limit expression is a product of two functions. Therefore we can distribute first the limit expression to the two factors and we evaluate the limit of each factor using the limit theorems that we know so far. Executing the solution:
ø ùø ù û ýû xxxxx 44444 x x xx 9lim2lim3lim4lim9234lim ý
ûø ùø ù ýûø ùø ù xxxxxx 444444 x x 9limlim2lim3limlim4lim ý
ûø ùø ù ýûø ùø ù ý û ýû ý û ýû ý 11998316942344
x 4 ø ùø xx ù ü 199234lim
Alternative Solution : We can combine the concepts of limit theorems 3 and 4 in a single line of a solution by distributing the limit expression to each factor and each term. This is one way to reduce the length of our solution of evaluating limits. Doing this:
ø ùø ù ûø ùø ù ýûø ùø ù xxxxxxx 4444444 x x xx 9limlim2lim3limlim4lim9234lim ý
ûø ùø ù ýûø ùø ù ý û ýû ý û ýû ý 11998316942344
x 4 ø ùø xx ù ü 199234lim
The next limit theorem will be the one that we will use if the variables involved have exponents in it.
Limit Theorem # 5: The Limit of the n th Power of a Function
If lim ax ø ù Lxf and n is any real number, then:
lim ax û ýø ù n Lxf n
To apply limit theorem # 5 correctly, we evaluate the limit of the base of the exponential expression by using the other limit theorems that we know so far, and simplify. That is:
ax û ýø ù n ûlimlim ax ø ù xfxf ý n
Ln
Example 9 : Evaluate x 3 ø 3 xx 432lim ù
The first term involves a variable having an exponent of 3. So we can apply here the concept of limit theorem # 5. We will show two possible solutions here: one is the step by step applications of the limit theorems , and the other is the combination of concepts of limit theorems 3, 4 and 5 to shorten our solution.
Solution 1 : 3 ø 3 ù 3 3 x x xx xxxx 33 4lim3lim2lim432lim
ø ùø 33 3 ù ø ùø ù xxxxx 333 x x 4limlim3limlim2lim
Example 10 : Evaluate x 3 x 2 166lim
First step here is we apply limit theorem number 6, since we are involved with a limit having a radical symbol. We next evaluate the limit of the expression inside the radical symbol by using the previous limit theorems that we know. Executing the solution:
3 2 xx 3 ø xx 2 166lim166lim ù
ø ùø ù ø ùø ù 2 ø ùø ù 16541696163616limlim6lim
3
2 33
x xxx
x 3 x 2 ü 38166lim
Since 38 is already a simplified radical, we leave this as our final answer. For instances that the radical obtained can still be simplified (e. g. 24 , 54 , 381 ), we are required to transform them to a simplified radical form by extracting roots of suitable factors of the radicand.
Note also that we directly distributed the limit expressions to each terms and factors. We can also present our solution by using the limit theorems one by one, for each line. But it is better as a suggestion that we acquaint ourselves with this shorter method of evaluating limits.
The last limit theorem for this section will be involving the limit if we are involved with a quotient of two functions.
Limit Theorem # 7: The Limit of the Quotient of Two Functions
####### If lim ax ø ù Lxf and lim ax ø ù Mxg , then:
####### ø ù
####### ø ù M
L xg
xf lim ax ; 0M
This limit theorem is used if the limit of the function involved has a numerator and a denominator , provided that the limit of the denominator is not zero ( 0M ). If we found out at first hand that the limit of the denominator is 0, then we cannot apply this limit theorem. As we go along with our subject, we will illustrate one by one on the possible procedure/s that we will apply if the denominator is zero. But provided that the limit of the denominator is not
zero, we just distribute the limit expression to the numerator and denominator and find each of the limits by using the previous limit theorems. In general, to illustrate:
ø ù
ø ù
ø ù
ø ù xg
xf xg
xf ax
ax ax
lim
lim lim
M
L ; 0M
What if the limit of the denominator is 0? Recall that in this case, limit theorem # 7 cannot be used directly. But we can still use this limit theorem after redefining the function that is given. Redefining the function can be done by doing one or more of these possible manipulations: (a) by factoring , (b) by rationalizing the radical in the numerator or (c) by simplifying the function using techniques of algebra. Applying one or more of these possible manipulations can help us apply the concept of limit theorem # 7. We will illustrate all of these cases one by one as we proceed with examples.
Example 11 : Evaluate 17
lim 4 x x x
Before trying to apply limit theorem # 7, we must check first if the limit of the denominator is not 0. This is because the moment the limit of the denominator is zero, the concept of the limit of a quotient cannot be applied yet. To compute for the limit of the denominator (we can also do this mentally, if we can):
ø ù û ø ùýø xx ù xxxx 4444 ø ùø ù 271281471limlim7lim17lim ; 027
Since the limit of the denominator is not 0, we can apply limit theorem # 7 directly. Distributing the limit expression to the numerator and denominator, and evaluating each limit, we have:
)17(lim
lim 17
lim 4
4 4
x
x x
x x
x x
ø ùø ù 128
4 147
4 1lim)lim)](7(lim[
lim
444
4
xxx
x x
x
27
4 17
lim 4
ü x
x x
####### ø ùø ù
5
lim 5 5
lim 25 55
z z5z z
z z
2 z
####### z lim 5 ø ù 5-z
The resulting new function if you may notice has no denominator anymore, and we can now apply the limit theorems that we know. Doing this:
####### ø ù 555limlimlim
5
lim 25 5555
zzz
2 z z z5-z z
0 5
lim 25 5 ü z z 2 z
In general, the method of factoring can be used if either the numerator or denominator can be factored by using a certain method. Therefore, at this point, we must recall all possible methods of factoring polynomials, from your junior high school or senior high school mathematics. This technique can often be successful if common factors in the numerator and denominator cancel out.
Example 14 : Evaluate 1
lim 1 1
s s 3 s
The limit expression involved is in quotient form where the limit of the denominator is 0. If this is the case where the limit of the denominator is zero, we will think if either the way of factoring, rationalization or simplifying the function is applicable. Being observant in the involved function, note that the numerator can be factored using the concept of a sum or difference of two cubes. Recall from algebra that a sum or difference of two perfect cubes can be factored in this form:
3 ø ùø babababa 223 ù
Therefore, the numerator s 3 1 can be factored as:
3 ø ùø 2 ssss 111 ù
Replacing the given numerator of the limit expression by its factored form and evaluating the resulting limit, we have:
ø ùø ù
1
lim 11 1
lim 12 11
s
sss s
s s
3 s
1 ø 2 ù ø 1 ù 2 ssss ssss 11 1limlimlim1lim
####### ø ù 111111
1
lim 12 1 s s 3 s
3 1
1
ü s
s 3
Example 15 : Evaluate 4
lim 2 4
x x x
The involved function still is in fraction form where the limit of the denominator is zero. Recall that if we encounter this case, we analyze the given function if we can factor, rationalize or simplify. Neither the numerator nor the denominator is factorable; therefore the method employed in the last two examples is not applicable.
But upon noticing that the numerator has a presence of radical symbols, we can try to rationalize the numerator by multiplying both the numerator and denominator by the conjugate of the denominator. Then we simplify for the limit of the resulting function to be possibly evaluated using the limit theorems that we know. This method is introduced in the previous section involving evaluating functions.
Recalling that the conjugate of x 2 is x 2 , we multiply this conjugate to both numerator and simplify. Executing the process:
)2)(4(
lim 4 2
2 4
lim 2 4
lim 2 444
xx x x
x x
x x
x xxx
2
lim 1 x 4 x ;
22
1 24
1 2limlim
1lim
44
4
xx
x x
4
1 4
lim 2 4 ü x x x
Example 16 : Evaluate h
2h h lim 2 0
Based on the given, the limit of the denominator is still 0. To check if one of the three possible manipulations is possible, the method of factoring is not applicable. But upon
Note here that the limit of the denominator here is not zero anymore.
####### ø ù
4
1 4
lim 4 4
4
11 lim 44
xx
x x
x xx
####### ø ù
)4)(4(
1 )(lim)4lim(
1lim 4
lim 1 44
4 4
x x x x
x x
16
1 4
4
11 lim 4
ü x
x x
Example 18 : Evaluate ÷÷ ø
ö ÷÷ø
ö 9
1 3
lim 11 x 0 x x
From the first factor of the given function, note that we cannot evaluate the limits directly because the denominator will be zero. Therefore we must think if what possible manipulation (factor, rationalize or simplify) can be applied in the given. Sometimes, experimentation of the possible manipulations can be done on the given for us to evaluate its limit. One possible first step is we first simplify the given by expressing the two terms in the 2nd factor in terms of a common denominator. Doing this, rationalizing the numerator and evaluating the limits, we have:
÷
÷ ø
ö ÷
÷ ø
ö
÷÷ø
ö ÷÷ø
ö
93
lim 391 9
1 3
lim 11 00 x
x x x x
xx
÷÷ ø
ö ÷
÷ ø
ö
39
39 93
lim 391 0 x
x x
x x x
####### ø ù
ú
ú û
ù ú
ú û
ù
ú
ú û
ù ú
ú û
ù
)39( lim 1 )39(
lim 991 00 xx
x x xx
x x
xx
)39(
lim 1 x 0 xx
)390(
1 )3lim9limlim(9limlim)3lim(
1lim
000000
0
xxxxxx
x xx
)6(
1 )33)(3(
1 9
1 3
lim 11 0 ÷÷ø
ö ÷÷ø
ö x x x 54
1 9
1 3
lim 11 0 ÷÷ø
ö ÷÷ø
ö ü x x x
So far in our examples of evaluating limits of functions, we just focused on limits involving polynomial functions. On our future sections, we will have the process of evaluating limits of some other functions, especially trigonometric functions. Also, the cases of limits of polynomial functions involving a quotient that we evaluate so far is that if the denominator is zero, we apply any of the three possible manipulations, that is either we factor, rationalize or simplify. In general, there are still other limit expressions of polynomial functions where none of the three manipulations are applicable. We will proceed to those cases in our next sections.
In this last example, we will show on how to evaluate limits of functions if we are given with the graph of that function.
####### Example 19 : The domain of ø ù xf is
####### û ý 5,5. (a) What are the values of f ø ù 4 ,
####### f ø ù 3 , f ø ù 3 and f ø ù 4? (b) x lim 4 ø ù xf ,
####### x lim 3 ø ù xf , x lim 3 ø ù xf , x lim 4 ø ù xf?
####### In this example, we are to find expressions for function values of ø ù xf and limits of
functions for certain values of x. This will emphasize further the difference between the function value and the limit. Take note that both the function value and the limit of a function represent the value of a y -coordinate.
####### To emphasize the difference, let us have f ø ù 4
####### and x lim 4 ø ù xf. Recall that f ø ù 4 represents the
actual value of y if the value of x is -4. Referring to the graph, if we are to find the actual value, then we must trace on the graph the actual point (solid dot) and determine there the value of y where x = -4. Note that at x = -4, the actual point involved in the function is at (-4, 6). Therefore,
####### f ø ù 64. Doing this similarly to other function
values that we are to find (points involved are encircled in the figure), we have:
####### a. f ø ù 64 ; f ø ù 43 ; f ø ù 73 ; f ø ù 34
####### Now, how about x lim 4 ø ù xf? To evaluate the limit of a function given its graph, we
consider the main curve / graph involved (semicircle). Given this main curve, we note
Test Your Understanding : Evaluate the following limits analytically, if they exist.
####### 1. x 5 ø x 73lim ù 5.
53
lim 12 12
xx x x
2. z 2 ø z 3 8lim ù 6.
3
lim 18 1
r r r
3. x 2 ø 2 xx 12lim ù 7.
12
lim 43 2
2 4
xx xx x
- 15
lim 54 3
x x x 8.
3 35 lim 25 x
x x
EXERCISE 1 :
A. Evaluate the following limits analytically, if they exist.
- 7
lim 249 7
x x x (Ans: 14) 7. 62 lim 5 3
2 2
t t t
- 32
lim 294 2/1
x x x (Ans: -6) 8. 5 lim 225 5
z z z
- 492
lim 1683 2
2 4
ss ss s (Ans: 16/7) 9. 36254 lim 20173 2
2 4
xx xx x
- 2
lim 38 2
y y y (Ans: 12) 10. 94
lim 278 2
3 2/3
t t t
- 372
lim 9 3
2 3
yy y y (Ans: 5
30 ) 11. 1
lim 25 1
x x x
- 1
lim 1 1
x x x (Ans: 1/2) 12. 6
6
11 lim 6
z z z
####### B. The domain of ø ù xf is ø 2, ý.
####### a. Determine the values of f ø ù 1 , f ø ù 0 , f ø ù 1 and
f ø ù 3?
####### b. Determine the values of x lim 1 ø ù xf , x lim 0 ø ù xf ,
####### x lim 1 ø ù xf , ø ù xf
x 3
lim
?
1 - limits notes
Course: Industrial Engineering (ERGO1)
University: University of the Assumption
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