14.1 Limits of Functions:
We have studied the behavior of a function at a particular value now:
LIMITS = deal with the behavior of the function near x=c rather than at x=c
Ex. Graph: f(x) = .1x4 - .8x³ + 1.6x² + 2x – 8
x-4
using window 0≤x≤8 and 0≤y≤3
and describe the behavior of f(x) (the function) when x is very close to 4
a) not defined at x = 4 (hole)
b) look at viewing window 3.5≤x≤4.5 and 0≤y≤3 and see what happens to the
values of f(x) when x is very close to 4  (use trace feature, look at table)
**Shows that as x gets closer and closer to 4 from either side of the graph, the
corresponding values of f(x) (the function) get closer and closer to 2.
 If carry out more decimals – you can make as close to 2 as you want
 Ex. f(3.999999) = 1.999984
f(4.00001)=2.000016
In words this is saying “ the limit of f(x) (the function) as x approaches 4 is 2.
Symbolically this says: lim f(x) = 2
x4
Informal definition of LIMIT
** let f be a function and let c be a real # such that f(x) is defined for all values of
x near x=c, except possibly at x=c itself.
a) Suppose that  whenever x takes values closer and closer but not = to c (on
both sides of c) the corresponding values of f(x)(the function) get very close
to, and possibly equal to the same real # L(limit)
b) and that the values of f(x)(the function) can be made arbitrarily close to L by
taking values of x close enough to c but ≠ c, then it is said that “the limit of
the function f(x) as x approaches c, is the #L and written lim f(x) = L
xc
Ex. f(x) = sin3x
x
a)
b)
c)
d)
find lim f(x)
x0
graph in window -2π≤x≤2π and -2≤y≤4
see that from left as x approaches 0 f(x) approaches 3
see that from right as x approaches 0 f(x) approaches 3
so limit = 3
Ex. find lim f(x) where f is the function defined by the following rules:
x-1
f(x) = 1 if x is an integer
-2 if x is not an integer
** can’t do this on calculator – must graph on your own
--when x is a # very close, but ≠ -1 (either greater than -1 or less than -1) the
corresponding value of f(x) is -2 and this is true no matter how close x is to -1,
So lim f(x) =-2
x-1
Limits and Function Values:
**If the limit of a function f as x approaches c exists, this limit may not be equal to f(c).
In fact, f(c) may not even be defined.
** Often the limit of a function as x approaches a point is equal to the value of the
function at that point:
Ex. if f(x) = -cos²πx – sinπx find f(-1) and lim f(x)
x-1
f(-1) = -cos²(-π) – sin(-π) = -(-1)² - 0 = -1
graph and then trace  as x approaches -1 from left and right, the function
approaches -1 so L=-1 (use window -3<x<3 and -2<y<2)
Non Existence of Limits – can fail to exist for several reasons:
#1. f(x) becomes infinitely large or infinitely small as x approaches c from either side
Ex. lim -1
x0
x²
graph this:
**as x approaches 0 from the left of the right, the values of f(x) get smaller and
smaller without bound – rather than approaching 1 particular number (look at
trace) so LIMIT DOES NOT EXIST
#2. F(x) approaches L as x approaches c from the right and f(x) approaches M with M≠L
As x approaches c from the left.
Ex. find lim -|2x| if it exists
x0
x
is undefined for x=0, and there are 2 possibilities
a) if x>0 then f(x) = -|2x| = - 2x = -2
x
x
b) if x<0 then f(x) = -(-2x) = 2x = 2
x
x
**as x approaches 0 from both sides of 0, the corresponding values of f(x) do not
approach the same real# as required by the definition of a limit, so DOES NOT
EXIST
#3. f(x) oscillates infinitely many times between 2 #’s as x approaches c from either side
Ex. find lim cosπ if it exists
x0
x
**the graph oscillates infinitely between -1 and 1 with waves becoming more and
More compressed as x approaches 0
** as x approaches 0, the function takes every value between -1 and 1 infinitely
Many times, so f(x) does not approach 1 particular #. Therefore limit
DOES NOT EXIST.
14.2 Properties of Limits
*Most functions that appear in calculus are combinations of simpler functions
2 easy but important cases where the limit of a function may be found by evaluating
the function:
#1. Limit of a Constant – if d is a constant, then lim d = d
xc
Ex. lim 5 = 5
x3
(as x approaches 3, the corresponding value of f(x) is always 5)
#2. Limit of the Identity Function – for every real #c, lim x=c
xc
Properties of Limits
**If f and g are functions and c,L, and M are numbers such that:
lim f(x) = L and lim g(x) = M
then:
xc
xc
#1. lim (f+g)(x) = lim [f(x) + g(x)] = L + M
xc
xc
#2. lim (f-g)(x) = lim [f(x) - g(x)] = L - M
xc
xc
#3. lim (f·g)(x) = lim [f(x) · g(x)] = L · M
xc
xc
#4. lim (f/g)(x) = lim [f(x)/g(x)] = L/M (for M≠0)
xc
xc
#5. lim√f(x) = √L (provided f(x)≥0 for all x near c)
xc
Limits of Polynomial Functions – solve using the properties
Ex. if f(x) = x² + 3x – 6 and lim f(x) = lim (x² + 3x – 6)
x-2
= lim (x²) + lim 3x – lim 6
x-2
x-2
x-2
(used properties 1 and 2)
x-2
= lim (x) · lim(x) + lim 3 · lim x – lim 6
x-2
x-2
x-2
x-2
= (-2 · -2) + (3· -2) – 6 = -8
(used properties 3)
x-2
(used limit of a constant and limit of x)
Limits of Polynomial Functions – if f(x) is a polynomial function and c is any real #,
then
lim f(x) = f(c)
xc
*In other words, the limit is the value of the polynomial function f at x=c
Limits of Rational Functions - let f(x) be a rational function and let c be a real # such
that F(c) is defined. Then lim f(x)=f(c).
xc
Ex. if f(x) = x³ + x² - 8x -12 find lim f(x)
x-4
x² - x -2
**graph it and see that the graph of f near x = -4 suggests that the limit is a # near
-1.56
** since f(x) is a quotient of the 2 functions, as x approaches -4, the limit can be
Found by evaluating the functions at x = -4 so
Lim x³ + x² - 8x -12
x-4
x² - x -2
= (-4)³ + (-4)² - 8(-4) – 12
(-4)² - (-4) – 2
= 28 = -1.556
18
Rational Function Limits  say that the limit of a rational function as x approaches c is
the value of the function at x=c  if the function is defined
there.
**when a rational function is not defined at a #, different techniques must be used
to find its limit there  if it exists
Ex. if f(x) = x² - 2x – 8 find lim f(x)
x-2
x+2
* plug in (-2) and will see that f(x) is not defined when x = -2 so the limit can’t be
found this way.
(x+2)(x-4) = x-4 for all x≠-2
(x+2)
--the definition of a limit = as x approaches -2, involves only the behavior
of the function near x=-2 and not at x=-2. Both f(x) and g(x) = x-4
have exactly the same values except at x=-2, so they must have the
same limit as x approaches -2
**so step 1: reduce the rational expression:
step 2: lim (x-4)  -2 -4 = -6
(as illustrated also by the calculator)
x-2
LIMIT THEOREM: If f and g are functions that have limits as x approaches c and
f(x) = g(x) for all x≠c then:
lim f(x) = lim g(x)
xc
xc
**says that if 2 functions have identical behavior, except possibly at x=c,
they will have the same limit as x approaches c **
Recall the difference quotient of a function: f(x+h) – f(x)
h
**it can be evaluated for a specific value of x, say x=c to obtain a new form:
f(c+h) – f(c)
h
ex. if f(x) = 2x² find lim f(-3+h) – f(-3)
h 0
h
14.2A
One Sided Limits
*The limit of f(x) as x approaches 4 from the right is 2
lim f(x) = 2
x4+
+ = indicates only values of x with x>4 are considered
*The limit of f(x) as x approaches 4 from the left is -1
Lim f(x) = -1
x4-
-
= indicates only values of x with x<4 are considered
One Sided Limits
 different from 2 sided b/c f(x) does not approach a single #, so the limit
does not exist
1) lim f(x) = L if x>c  function doesn’t need to defined when x<c
xc+
2) lim f(x) = L if x<c  function doesn’t need to be defined when x>c
xc-
Ex. find: lim (√2-x – 3)
x2-
a) is defined only for x≤2 – that is x=2 and all values to the left of 2
(graph and look at the table)
b) the values of f(x) approach -3 as x approaches 2 from the left, so
lim (√2-x – 3) = -3
x2-
Computing 1-Sided Limits – all properties of limits, limits of polynomial functions, and
the limit theorem remain valid it “xc” is replaced if
either “xc+” or “xc-“
ex. find lim √16-x²
**(is defined only when -4≤x≤4, so compute the
limit as x approaches -4 from the right by using
properties
x-4+
=√lim(16-x²) = √lim 16 – lim x² = √lim 16 – (lim x)(lim x) = √16 – (4)(4) = 0
x-4+
x-4+
x-4+
x-4+
x4+
x-4+
3 Types of Limits:
1. left hand
2. right hand
3. “two-sided”
2-sided limits : let f be a function and let c and L be real #’s, then lim f(x) = L exactly
xc
when:
lim f(x) = L
and lim f(x) = L
xc-
Ex. From the graph below find the following limits:
a) lim f(x) =
x5
xc+
b) lim f(x) =
x5-
c) lim f(x) =
x5+
d) lim f(x) =
x 3-
e) lim f(x) =
x3+
14.4 Continuity
Informally- let c be a real # in the domain of f
**the function is continuous at x=c if you can draw the graph of f at and near the
point (c,f(c)) without lifting your pencil from the paper  if the graph at (f,f(c)) is
connected and unbroken.
Vs. discontinuous = not continuous if the graph has a break, gap or hole, or jump
at x=c
*Calculators – may present misleading info. about continuity b/c it plots a # of points
then connects them with line segments to produce a curve – sort of
assumes it is continuous
2 Possibilities of Continuity:
1. Continuity at an interior point – if that is the case then 2 statements are true:
a) f(x) must be defined for x=c
b) f(x) must be defined for x=t, when t is any # near c  if not defined for t
near c, then there is a hole and pencil would
have to be lifted.

Figure:
If f(c) is defined, there are 2 conditions that can prevent a function from being
Continuous at x =c
1) There is a jump at x=c  the limit of f(x) as x approaches c doesn’t exist
2) There is a hole in the graph at x=c, that is  the limit of f(x) exists at x=c
but is not = f(c)
Formal Definition of Continuity = let f be a function that is defined for all x in some
open Interval containing c. Then f is said to be continuous at x=c, under the following:
1) f(c) is defined
2) lim f(x) exists
xc
3) lim f(x) = f(c)
xc
**look at figure 14-4, does it match all 3 parts of definition?
Ex. without graphing show that the function f(x) = √2x(2-x) is continuous at x=3
x²
Step 1: show that f(3) = √2x(2-x)
x²
f(3) = √2(3)(2-3) = -√6
3²
9
Step 2: lim f(x) = lim√2x(2-x) = -√6
x3
x²
9
**since lim f(x) = f(3) then f is continuous at x =3
x3
Continuity of Special Functions
1. Every polynomial function is continuous at every real # ex: x³-4x² + 1
2. Every rational function is continuous at every real # in its domain ex. (x²-5)/(x-2)
3. Every exponential function is continuous at every real # ex. y = 34x
4. Every logarithmic function is continuous at every positive real # ex. y=3logx
5. f(x) = sinx and g(x) = cosx are continuous at every real # ex. y=sinx
6. h(x) = tanx is continuous at every real # in its domain ex. y=tan x
(continuous except at odd mults of π/2)
Continuity on an Interval:
#1. Continuity from the right – a function is continuous from the right at x=a provided
that lim f(x) = f(a)
(a,f(a))
xa+
#2. Continuity from the left – a function is continuous from the left at x=b provided that
lim f(x) = f(b)
(b,f(b))
xb-
Ex. show that f(x) = √3-x is continuous from the left at x=3
a) f(x) = √3-x is not defined for x values that make 3-x<0  x>3
b) so f(x) = √3-x is not defined for x>3
f(3) = √3-3 = 0
So…
lim √3-x  √3-3 = 0
x3-
Summary of Continuity on the Interval:
1) A function f is said to be continuous on an open interval (a,b) provided that f is
continuous at every value in the interval.
2) A function f is said to be continuous on a closed interval [a,b] provided that f id
continuous from the right at x=a, and continuous from the left at z=b, and continuous at
every value in the open interval (a,b)
Ex. discuss the continuity of the function:
Discontinuous at x = -1 and x = 2
Continuous at [-4,-1], (-1,2), [2,∞)
Properties of Continuous Functions:
**if the functions f and g are continuous at x=c, then each of the following functions is
also continuous at x=c
1)
2)
3)
4)
The sum function f+g
The difference function f-g
The product function fg
The quotient function f/g g(c)≠0
Ex. assume that f(x) = cos x and g(x) = x4 – 2x² + 4 are continuous at x=1. Prove the
following are continuous at x=1
a) cosx – (x4 – 2x² + 4)  = (f-g)(x) difference of a continuous function
b) (cosx)( x4 – 2x² + 4)  = (fg)(x) product of a continuous function
Removable Discontinuity – when a function is not defined at x=c
**basically if you have a hole – a rational function in which a factor can cancel
Non-removable Discontinuity – an asymptote
Ex. f(x) = x2-1
x -1
** this has a removable discontinuity at x = 1
Ex. f(x) = 1
x
** this has a non-removable discontinuity at x = 0
Composite Functions – composition of functions often used to construct new functions
from given ones.
1) Continuity of Composite Functions = if the function f is continuous at x=c and the
function g is continuous at x = f(c), then the composite function g◦f is continuous
at x=c
The Intermediate Value Theorem – if the function f is continuous on the closed interval
[a,b] and k is any # between f(a) and f(b) then there exists at least
One #c between a and b such that f(c)=k
 Assume f is continuous on [a,b]
 Choose any # k on the y-axis between f(a) and f(b)
 The Intermediate Value Theorem guarantees that any horizontal line through k
will intersect the graph
 From that point of intersection, move down vertically to the x-axis to get a #c
guaranteed by the theorem
**The theorem guarantees the existence of at least 1#c, but there may be more than 1
**If f is continuous on the interval [a,b] and f(a) and f(b) have opposite signs, then 0 is a
# between f(a) and f(b) and 0 is a solution to the equation
14-5 Limits Involving Infinity
Infinity – generally speaking indicates a situation in which some numerical quantity gets
larger and larger without bound  so can be made larger than any given #
Negative infinity – indicates a situation in which some numerical quantity gets smaller
and smaller without bound  so can be made smaller than any given #
** as x approaches 3 from the left and right, the corresponding values of f(x) get larger
and larger without bound say lim = ∞
x3
**as x approaches 1 from the left and right, we say lim = -∞
x1
** find lim
x5-
and
lim
x5+
Ex. describe the behavior of f(x) = 9 near x = 0
X4
**values of f(x) – using trace on calc. get large w/o bound
as x approaches 0 from left or right
** lim 9 =
x0 4
x
Ex. describe the behavior of g(x) = -5 near x = 3
x²-x-6
**graph shows not continuous at x = 3
**to the left of x=3 values of g(x) get large w/o bound
so lim =
x3-
**to the right of x=3 values of g(x) get small w/o bound
so lim =
x3+
Vertical Asymptotes: the vertical line x=c is a vertical asymptote of the graph of the
Function f if at least 1 of the following is true:
lim f(x) = ∞
lim f(x) = ∞
lim f(x) = ∞
xc-
xc+
xc
lim f(x) = -∞
lim f(x) = -∞
lim f(x) = -∞
xc-
xc+
xc
Limits at Infinity – let f be a function that is defined for all x>a for some #a
(the end behavior when x takes very large or small values)
If: 1) as x takes larger and larger positive values, increasing without bound, the
corresponding values of f(x) get very close, and possibly are = to a single
real # L
2) the values of f(x) can be made arbitrarily close (as close as you want) to L
By taking large enough values of x
Then: the limit of f(x) as x approaches infinity is L written as: lim f(x) = L
x∞
** Limits as x approaches infinity or negative infinity correspond to horizontal
asymptotes
Horizontal Asymptotes: the line y = L is a horizontal asymptote of the graph of the
function f if either:
lim f(x) = L
x∞
or
lim f(x) = L
x-∞
Ex. Describe the behavior of f(x) = 1 as x approaches ∞ and approaches -∞
x-3
**when x is very large + #, 1 is a positive # close to 0
x-3
**when x is very small - #, 1 is a negative # close to 0
x-3
**suggests that y = 0 is a horizontal asymptote b/c
lim = 0 lim = 0
x∞
x-∞
Ex. Describe the behavior of f(x) = -x3 + 9x -7 as x approaches ∞ and -∞
**No polynomial graph has a horizontal asymptote
--no poly. function has a limit as x approaches ∞ or -∞
Limit of a Constant function: If c is a constant then:
lim c = c
and
lim c = c
x∞
x-∞
Properties of Limits at Infinity : If f and g are functions and L and M are #’s such that:
lim f(x) = L
and
lim g(x) = M
then:
x∞
x∞
#1. lim (f+g)(x) = lim [f(x) + g(x)] = L + M
x∞
x∞
#2. lim (f-g)(x) = lim [f(x) - g(x)] = L - M
x∞
x∞
#3. lim (f·g)(x) = lim [f(x) · g(x)] = L · M
x∞
x∞
#4. lim (f/g)(x) = lim [f(x)/g(x)] = L/M (for M≠0)
x∞
x∞
#5. lim√f(x) = √L (provided f(x)≥0 for all x near c)
x∞
Limit Theorem: if c is a constant, then for each positive integer n
Lim c = 0
x∞ xn
lim c = 0
x-∞ xn
** used to determine the limit, if it exists, of any rational function as x
approaches ∞ and -∞
Ex. describe the end behavior of f(x) = -4x² + 2x -5 then justify your conclusion:
5x² + 6x -9
Step 1: take lim and divide by highest degree of x
x∞
Key Points for Rational Functions:
#1. if degree of numerator < degree of denominator
LIMIT = 0
#2. if degree of numerator = degree of denominator
LIMIT = ratio of highest
Power terms
#3. If degree of numerator > degree of denominator
LIMIT = does not exist
Limits of rational functions with square roots: