1. LIMIT OF A FUNCTION AND LIMIT LAWS
1.1 What is “Limit”
Frequently, when studying a function y=f ( x ) , we find ourselves interested in the
function’s behavior near a particular pointxo , but not at xo itself. This is due to various
reasons. One of the reasons could be because when trying to evaluate a function at xo , it leads to division by zero, which is undefined! Let’s have a look at the following example:
Example 1
Observe the behavior of the following function nearx=1 .
f ( x )= x2−1x−1
Solution:
Notice from graph 1 that there is actually a ‘hole’ atx=1 . Evaluating f at x=1 gives:
f (1)=12−11−1
=00
But ( 0
0)is “indeterminate”, meaning, we
can’t determine its value. So, f (1) is not defined, that is why here is a hole in graph 1. Graph 1
Instead of evaluating f directly atx=1 , let’s observe the value of f ( x ) when we approachx=1 :
x 0.9999 0.999 0.9 0.5 1 1.0001 1.001 1.01 1.1 1.5 f (x) 1.99990 1.99900 1.90000 1.50000 ? 2.00001 2.00100 2.0100
0 2.100 2.5
As seen from the table above, the closer x gets to 1, the closer f ( x ) seems to get to 2.
Let’s generalize the idea illustrated in Example 1:
Limit (informal definition): If the values of f ( x )can be made as close as we like to L by
taking values of x sufficiently close to xo (but not equal toxo ), then we write
limx→x o
f ( x )=L,
Right sideLeft side
which is read “the limit off ( x ) as x approaches xo is L.”
For instance, in Example 1, we would say that f ( x ) approaches the limit 2 as x approaches 1 and write:
limx→1
f ( x )=2or
limx→1
x2−1x−1
=2
Remember, to findlimx→x o
f ( x ), we should be able to approach as close as we like toxo . This
means: the function f should be defined “everywhere nearxo ”; otherwise, we would say: the limit does not exist. Let’s have a look at few more examples.
Graph 2: limit exist because x can approach 2 as close as we like
Graph 3: limit does not exist because x cannot approach 2 as close as we like
1.2 One -sided Limits: what is right hand limit & left hand limit?
1. If the value of f ( x ) approaches the number L as x approaches c from the right,
we writelimx→c+
f (x )=L(one-sided limit – right-hand limit)
or f ( x )→ L as x→c+
2. If the value of f ( x ) approaches the number M as x approaches c from the left,
we writelimx→c−
f (x )=M(one-sided limit – left-hand limit)
Theorem A function f ( x ) has a limit as x approaches c if and only if it has left-hand and right –hand limits and these one-sided limits are equal:
xx
f(x)
f(x)
x x
f(x)
f(x)
limx→ c
f ( x )=Lif and only if
limx→c−
f (x )=limx →c+
f ( x )=L
If both one-sided limits do not have the same value then limx→ c
f ( x )does not exist.
Note: The limit of a function f ( x ) as x approaches c does not depend on the value of the function at c [limit does not depend on f(c)]
Example 2:
Graph 4
At x=1 :limx→1−
f ( x )=0 even though f (1)=1 ,
limx→1+
f ( x )=1,
limx→1
f ( x ) does not exist. The right and left hand limits are not equal.
At x=2 :limx→2−
f (x )=1,
limx→2+
f (x )=1,
limx→2
f ( x )=1 even though f (2)=2 ,
1.3 The Limit Laws
Theorem:
If limx→ c
f ( x )=L , limx→c
g( x )=M both exist, and k is a constant, then
a)limx→ c
[ f ( x )+g ( x ) ]= limx→ c
f ( x )+limx→c
g( x )=L+M
b)limx→ c
[ f ( x )−g( x ) ]=limx→c
f ( x )−limx→c
g ( x )=L−M
c)limx→ c
[ f ( x ). g( x ) ]= limx→c
f ( x ) . limx→ c
g( x )=LM
d)limx→ c
kf ( x )=k limx→c
f (x )=k⋅L
e)
limx→ c
f ( x )g ( x )
=limx→c
f ( x )
limx→c
g ( x )=
LM , M≠0
f)limx→ c
s√ (f (x ))r= s√( limx→cf ( x ))r=
s√Lr=Lr
s
g)limx→ c
[ f (x )]n=[ limx→ cf ( x )]n=( L )n
If f is the identity function f ( x )=x , then for any value of xo
⇒ limx→ xo
f ( x )= limx→ xo
x=x0
If f is the constant function f ( x )=k , then for any value of xo
⇒ limx→ xo
f ( x )= limx→ xo
k=k
1.4 Computing Limits
There are various algebraic methods to solve for limits but the first step would always be substitution. If substitution results in
1. A real number, negative infinity (−∞ ), positive infinity (+∞ ) → limit is found
2. Zero over infinity0
±∞ → answer to the limit is zero.
3. A real number over zero
30 or infinity over zero
±∞0 → Limit does not exist.
Example 3
Evaluate the following limits.
A) limx→5
( x3−3 x+5 ) Substitute x with 5:
limx→5
( x3−3 x+5 )=(5)3−3 (5 )+5=115
B) limx→2
5 x3+4x−3 =
limx→2
(5 x3+4 )
limx→2
( x−3 )=−44
C) lim
x→−2
3 x+4x+5
=−6+4−2+5
=−23
D) limx→2
1x−2=
limx→2
(1)
limx→2
( x−2 )=
10=?
use the sign graph to determine whether the answer is +∞ ,−∞ or both.
⇒ limx→2
1x−2 does not exist. (+∞ from the right, and −∞ from the left)
E) limx→3
f ( x ) for
f ( x )={ x2−5 , x≤3√ x+13 , x>3
From the left,limx→3−
f ( x )= limx→3−
( x2−5 )=4
From the right,limx→3+
f ( x )= limx→3+
√x+13=4
Both one-sided limits are the same⇒ lim
x→ 3f ( x )=4
However, if substitution results in indeterminate value such as ( 0
0), then we have to convert
the function into a suitable form where substitution can give one of the three results mentioned above. The following examples illustrate how to resolve indeterminacy by using algebraic methods.
Example 4 - Resolving indeterminate form of ( 0
0) by factorizing:
A) Evaluate limx→2
x2−4x−2
Substituting x with 2:
limx→2
22−42−2
=00⇒
indeterminate
Factorizing ( x2−4 ) into( x+2)( x−2 ):
=limx→2
( x+2)( x−2)( x−2)
Canceling ( x−2 )in numerator and denominator, and substituting x with 2:= lim
x→2( x+2 )=4
B) Evaluate limx→1
x3−14 x2−5 x+1
Substituting x with 1:
limx→1
13−14−5+1
=00⇒
indeterminate
Factorizing ( x3−1) into( x−1)( x2+x+1 )
and (4 x2−5 x+1) to ( x−1)( 4 x−1 )
=limx→1
( x−1 )( x2+x+1)( x−1 )(4 x−1)
Canceling ( x−1) in numerator and denominator, and substituting x with 1:
=limx→1
( x2+x+1 )( 4 x−1 )
=1
Example 5 - Resolving indeterminate form of ( 0
0) by conjugate multiplication:
A) Evaluate limt→0
√t 2+9−3t 2
Substitute t with 0:
limt→0
√02+9−302
=00⇒
indeterminate
Multiply with conjugate and simplify:
=limt→0
√t 2+9−3t 2
. √ t2+9+3√ t2+9+3
=limt →0
t2
t2(√ t2+9+3 )
Cancel t2
in numerator and denominator and substitute x with 0:
=limt →0
1( √t 2+9+3 )
=16
Theorem-(The Squeezing Theorem or Sandwich Theorem) Suppose that g( x )≤h( x )≤ f ( x )for all x in some open interval containing c (as shown in graph 5), except possibly at x = c itself. Suppose also that
limx→c
g( x )= limx→c
f ( x )=L
Then,limx→ c
h( x )=L
Graph 5 illustrating squeezing theorem
Squeezing theorem helps us to establish several important limit rules such as:
Theorem (limit rules):
limθ→0
sin θθ
=1 limθ→0
cosθ−1θ
=0
Proof:This proof is for the limit as θ→0+. The case for θ→0− can be proved in exactly the same manner.Consider the graph to the right. Notice that the area of the triangle OAP is less than the area of the sector OAP which is in turn less than the area of the triangle OAT. Let’s estimate each of these areas in turn.
Area of triangle OAP =
12
base×height=12(1)sin θ=1
2sin θ
Area of sector OAP =
12
r2θ=12(1)θ= θ
2
Area of triangle OAT =
12
base×height=12(1) tan θ=1
2tan θ
Using our initial observation, one has12
sin θ≤12
θ≤ 12
tan θ , 0≤θ≤ π2 (1)
From the left hand side of (1) one has sin θ≤θor
sin θθ
≤1if θ≥0 . From the right hand side
we have θ≤tan θ or cosθ≤sin θ
θ ifθ≥0 . In other words, cosθ≤sin θ
θ≤1
At this point, if we take the limit as θ→0 and apply squeezing theorem:
limθ→0
cosθ≤limθ→0
sin θθ
≤limθ →0
1
Since the left-hand side and right-hand side has the limit of 1, we can conclude that
limθ→0
sin θθ
=1
Example 7:
1.limx→0
tan xx
=limx→0 (sin x
x. 1cos x )=(1)(1)=1
2.limθ→0
sin 2 θθ
=limθ→ 0 (2⋅sin 2θ
2 θ )=2⋅( limθ→0
sin 2 θ2 θ )=2
3.limx→0
sin 3 xsin 5 x
=limx→0(sin 3 x
x÷sin 5 x
x )= limx→0 [(3⋅sin 3 x
3 x )÷(5⋅sin 5 x5 x )]=3
5
1.5 End behavior of function (limits asx→±∞ )
The behavior of a function toward the extremes of its domain is sometimes called its end behavior. In this section, we will use limits to investigate the end behavior of a function as x→±∞ . The symbol for infinity ( ∞ ) does not represent a real number. We use ∞ to describe the behavior of a function when the values in its domain or range outgrow all finite bounds.
Finite limits as x→±∞ (definition):
1. We say that f ( x )has the limit L as x approaches infinity and writelim
x→+∞f ( x )=L
2. We say that f ( x )has the limit L as x approaches minus infinity and writelim
x→−∞f ( x )=L
Geometrically, if f ( x )→Lasx→+∞ , then the graph of y=f ( x ) eventually gets closer and closer to the line y=L as the graph is traversed in the positive x-direction (see graph 6). And if f ( x )→Lasx→−∞ , then the graph of y= f ( x ) eventually gets closer and closer to
the line y=L as the graph is traversed in the negative x-direction (see graph 7). In either case, we call line y=L a horizontal asymptote of the graph of f.
Graph 6
Graph 7
horizontal asymptote (definition): A line y=b is a horizontal asymptote of the graph of a function y=f ( x ) if either
limx→∞
f ( x )=bor
limx→−∞
f ( x )=b
Let’s begin by obtaining the limits of some simple functions and then use these as building blocks for finding limits of more complicated functions.
Theorem: let k be a real number.
limx→+∞
k=k limx→−∞
k=k (constant function)lim
x→+∞x=+∞ lim
x→−∞x=−∞ (linear function)
limx→+∞
1x=0 lim
x→−∞
1x=0 (reciprocal function)
Theorem: If r>0 is a rational number such that xr
is defined for all x, then
limx→∞
1xr
=0and
limx→−∞
1xr
=0
The Limit Laws x→±∞
If lim
x→±∞f ( x )=L , lim
x→±∞g( x )=M
both exist, and k is a constant, then
a)lim
x→±∞[ f ( x )+g( x ) ]= lim
x→±∞f ( x )+ lim
x→±∞g( x )=L+M
b)lim
x→±∞[ f ( x )−g ( x )]= lim
x→±∞f ( x )− lim
x→±∞g( x )=L−M
c)lim
x→±∞[ f ( x ) . g( x )]= lim
x→±∞f ( x ). lim
x→±∞g (x )=LM
d)lim
x→±∞kf ( x )=k lim
x→±∞f ( x )=k⋅L
y
x
y=LHorizontal asymptote y
x
y=LHorizontal asymptote
e)
limx→±∞
f ( x )g( x )
=lim
x→±∞f ( x )
limx→±∞
g (x )=
LM , M≠0
f)lim
x→±∞
s√( f ( x ))r= s√( limx→±∞
f ( x ))r=s√Lr=L
rs
Limits of polynomials as x→±∞
limx→+∞
xn=+∞ , n=1,2,3 , .. . limx→−∞
xn={+∞ , n=2,4,6 ,. ..−∞ , n=1,3,5 ,. ..
Example 8:lim
x→+∞8 x8 =+∞ lim
x→−∞5 x5=−∞ lim
x→−∞−5 x8 =+∞
Note: The end behavior of a polynomial matches the end behavior of its highest degree term.
limx→+∞
(c0+c1 x+. . .cn xn)= limx→+∞
cn xn limx→−∞
(c0+c1 x+. .. cn xn )= limx→−∞
cn xn
Example 9: lim
x→+∞7 x5−4 x3+2 x−9= lim
x→+∞7 x5=−∞
limx→−∞
−4 x8+17 x3−5 x+1= limx→−∞
−4 x8=−∞
Limits of rational functions as x→±∞
There are two methods to find the limit of rational f ( x )=
n (x )g( x ) functions as x→±∞ :
Method 1: Use lim
x→+∞
1xn
= limx→−∞
1xn
=0
Example 10:
limx→−∞
4 x2−x2 x3−5
= limx→−∞
( 4x−1
x2 )
(2−5x3 )
= 02=0
Method 2: Use the end behavior of polynomial at numerator and denominator.
Example 11: lim
x→−∞
4 x2−x2x3−5
= limx→−∞
( 4 x2)(2 x3 )
= limx→−∞
2x=0
Limits involving radicals x→±∞
When a polynomial or rational function is below the radical, we can evaluate the limit of the function first and then take the radical.
Example 12:
limx→+∞
3√ 3 x+56 x−8
=3√ limx→+∞
3 x+56 x−8
=3√ 12
When radical appears either at numerator or denominator only, it would be useful to manipulate the function to powers of 1/x. This can be achieved in both cases by dividing the
numerator and denominator by |x| and using the fact that √ x2=|x|.
Example 13:
A)
limx→+∞
√x2+23 x−6
= limx→+∞
√x2+2/|x|(3 x−6 )/|x|
= limx→+∞
√ x2+2/√x2
(3 x−6 )/x= lim
x→+∞
√1+ 2x2
3−6x
=13
B)
limx→−∞
√ x2+23 x−6
= limx→−∞
√ x2+2/|x|(3 x−6 )/|x|
= limx→−∞
√x2+2/√x2
(3 x−6 )/(−x )= lim
x→−∞
√1+2x2
−3+6x
=−13
Another simpler method for finding the limit of functions involving radicals is to use the fact
thatm√xn=x
nm
.
Example 14:
limx→+∞
√x3+23√3 x5−6
= limx→+∞
x3
2
x5
3
=0
, the answer is zero because numerator degree is lower than denominator’s one.
1.6 Infinite Limits And Vertical Asymptotes
Infinite Limits (definition)
If the values of f ( x ) increase without bound (infinitely) as x→a+ or x→a−
,
we writelimx→a+
f ( x )=+∞or
limx→a−
f ( x )=+∞
If the values of f ( x ) decrease without bound (infinitely) as x→a+ or x→a−
,
we writelimx→a+
f ( x )=−∞or
limx→a−
f ( x )=−∞
If both one-sided limits are +∞ , then limx→a
f ( x )=+∞
If both one-sided limits are −∞ , then limx→a
f ( x )=−∞
Limits as x→±∞ can fail (or does not exist) when
a) the values of f ( x ) increase or decrease without bound.lim
x→+∞f ( x )=+∞ or lim
x→+∞f ( x )=−∞ or lim
x→−∞f ( x )=+∞ or lim
x→−∞f (x )=−∞
b) the graph of the function oscillates indefinitely, the values of f ( x ) does not
approach a fixed number, we say that lim
x→+∞f ( x )
and lim
x→−∞f ( x )
does not exist.
Vertical asymptote(definition): A line x=a is a vertical asymptote of the graph of a function y=f ( x ) if eitherlimx→a+
f ( x )=±∞or
limx→a−
f ( x )=±∞
Example15:
1.limx→2+
x−3x2−4
=limx →2+
x−3( x−2 ) ( x+2 )
=−∞
2.limx→2−
x−3x2−4
= limx →2−
x−3( x−2 ) ( x+2 )
=∞
3.limx→2
x−3x2−4
=limx→2
x−3( x−2 ) (x+2 ) does not exist.
2. CONTINUITY
Definition:Interior point : A function y=f ( x ) is continuous at an interior point c of its domain if limx→ c
f ( x )= f (c ).
Endpoint: A function y=f ( x ) is continuous at a left endpoint a or is continuous at a right endpoint b of its domain if
limx→a+
f ( x )=f (a ) or
limx→b−
f ( x )= f (b ), respectively.
Continuity TestA function f ( x ) is continuous at an interior point x=c of its domain if and only if it meets the following three conditions:
1. f (c ) exists [ c lies in the domain of f ]
2.limx→ c
f ( x ) exists [ f has a limit as x→c ]
3.limx→ c
f ( x )= f (c )[ the limit equals the function value ]
- If one or more of the conditions are not satisfied, then f is discontinuous at c, and c is a point of discontinuity of f.
- If f is continuous at (a,b) ⇒ f is continuous at all points of an open interval (a,b).- If f is continuous on (−∞ ,+∞)⇒ f is continuous everywhere. Types of discontinuity: Removable discontinuity, Infinite discontinuity, Jump discontinuity and oscillating discontinuity.
Continuous FunctionA function is continuous on an interval if and only if it is continuous at every point of the interval.
Theorem: (Properties of Continuous Functions) If the functions f and g are continuous at x=c , then the following combination are continuous at x=c :1. Sums: f +g 2. Differences: f −g 3. Products: f . g 4. Constant multiples: k⋅f , for any number k
5. Quotients:
fg , provided g(c )≠0
6. powers: fr
s, provided it is defined on an open interval containing
c, where r and s are integers.A) Polynomials
Any polynomial is continuous everywhere. limx→ c
p( x )=p (c )
B) Rational Functions
If P( x )and Q( x ) are polynomials, then the rational function
P( x )Q( x ) is continuous wherever it
is defined (Q( x )≠0 ). The function is discontinuous at the points where the denominator is zero.
Example:f ( x )= x2+1
x2+x−20
Denominator: x2+x−20=0 ⇒ x=4 ,−5⇒ f ( x ) is discontinuous at points x = –5 and x = 4.
C) Root FunctionsAny root function is continuous at every number in its domain.
D) Trigonometric FunctionsTrigonometric functions are continuous at every number in their domains.
Theorem:If c is any number in the natural domain of the stated trigonometric function, thenlimx→ c
sin x=sin c ; limx→c
cos x=cos c ; limx→c
tan x=tan c
limx→ c
csc x=cscc ; limx→ c
sec x=secc ; limx→c
cot x=cot c
Note: The functions sin x and cos x are continuous everywhere.
E) Composite Functions (g∘f )
Theorem: If f is continuous at c and g is continuous at f (c ) , then the composite g∘f is continuous at c
Theorem: If g is continuous at the point b and limx→ c
f ( x )=b, then
limx→ c
g[ f ( x ) ]=g(b )=g[ limx→c
f ( x )]
Examples:
1.limx→2
( x2+7 )2=[ limx→2
( x2+7 ) ]2=[ 11 ]2=121
2. limsin [ g (x )]=sin [ lim g( x ) ] limcos [ g ( x ) ]=cos [ lim g (x )]
e.g.limx→π
sin( x2
π+x )=sin[ limx→ π
x2
π+x ]=sin π2=1
--------------------------------------------End of Chapter 1----------------------------------------------
