3B MAS
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Limit of a FunctionGraphically the limiting value of a function f(x) as x gets closer and closer to a certain value (say 'a') is obtained by moving along the curve from both sides of 'a' as x moves toward 'a'.
The limiting value of f(x) as x gets closer and closer to 'a' is denoted by
x alim f (x)
Right/Left Hand LimitsAs x moves towards 'a' from right (left) hand side, the limiting value of f(x) is denoted by
)x(flimand)x(flimaxax
Limiting and Functional ValueIf both sides limits are equal,
Otherwise, does not exist.
Note that may not equal to f(a)
)x(flim)x(flim)x(flimaxaxax
)x(flimax
)x(flimax
Example 1Find the limit of the f(x) as x approaches a for the following functions.
(a)
a
The limit does not exist as the function is not defined 'near' a.
Example 1 (cont'd)
(b)
a
The limit does not exist as the left side limit is not the same as right side limit.
Evaluating LimitsIf f(x) is not broken at 'a', use direct substitution to evaluate its limit as x approaches 'a'
Otherwise, find the left side and right side limits and check if they are equal.
Example 2Evaluate the following limits if they exist.
(a) f(x) = 2x – 5 as x 1
f(x) is not broken at x = 1, so use direct sub.352)5x2(lim
1x
Example 2 (cont'd)(b) f(x) = ln x as x 0
f(0) is not defined. So consider limit from both sides.
But f(x) is not defined for x < 0.So the limit does not exist.
xlnlim0x
Example 2(c) f(x) = 1/(x – 2) as x 2
f(2) is not defined. So consider limit from both sides.
Since the left side limit does not equal to the right side limit, the limit of the f(x) as x approaches 2 does not exist.
)2x/(1limand)2x/(1lim2x2x
Example 2 (cont'd)(d) f(x) = (x – 1)/(x2 – 1) as x 1
f(x) = (x – 1)/(x + 1)(x – 1) = 1/(x + 1)1/(x + 1) is not broken at x = 1, so use direct sub.
5.01x
1xlim
21x
Limits to InfinityIf f(x) = x + c, f(x) as x (note that x is the dominant term)
If f(x) = 1/x, f(x) 0 as x If f(x) = ax2 + bx + c, ax2 is the dominant term as x
)x
c
x
ba(xcbxax
222
Example 3Find the limit of f(x) as x (if they exist) for:
4x5x
2x7x3)x(f)d(
1xx
4x2x)x(f)c(
3x2x
x71)x(f)b(
1x4
3x2)x(f)a(
2
23
2
2
2
03x2x
x71limSo
xasx
7
x
x7
3x2x
x71)x(f)b(
2
1
1x4
3x2limSo
xas2
1
x4
x2
1x4
3x2)x(f)a(
2x
22
x
Example 3 (cont'd)
)x(flimand)x(flimSo
xasx3x
x3
4x5x
2x7x3)x(f)d(
11xx
4x2xlimSo
xas1x
x
1xx
4x2x)x(f)c(
xx
2
3
2
23
2
2
x
2
2
2
2
Example 3 (cont'd)
Example 4Find the following limits.
x
x4tanlim)d(
x
xcos1lim)c(
x
xtanlim)b(
x
xsinlim)a(
0x
0x
0x
0x
x
C
B
AO r
Consider the relationship between the areas OAC, sector OAC , and OAB
0
sinlimx
xInvestigating
x
x
C
B
AO r
Area of OAC = r2 sin x / 2Area of sector OAC = r2 x / 2Area of OAB = r2 tan x / 2
Example 4 (cont'd)
x
C
B
AO r
Area of OAC = r2 sin x / 2Area of sector OAC = r2 x / 2Area of OAB = r2 tan x / 2
So (size of areas)
r2 sin x / 2 < r2 x / 2 < r2 tan x / 2 sin x < x < tan x1 < x / sin x < 1 / cos x1 > sin x / x > cos x
Take limit as x 0 to get
That means sin x x as x 0
1x
xsinlim
0x
4x4
x4tanlim4
x4
x4tanlim4
x
x4tanlim)d(
0
01xcos1
xsin
x
xsinlim
)xcos1(x
xcos1lim
xcos1
xcos1
x
xcos1lim
x
xcos1lim)c(
1xcos
1
x
xsinlim
x
xtanlim)b(
0x40x0x
0x
2
0x0x
0x
0x0x
Example 4 (cont'd)
ContinuityGraphically a graph is continuous at x = a if it is not broken (disconnected) at that point.
Algebraically the limit of the function from both sides of 'a' must equal to f(a).
)a(f)x(flim)x(flimaxax
Example 7Determine if the given function is continuous at the given point.
(a) f(x) = | x – 2 | at x = 2
(b) f(x) = x at x = 0
(c) f(x) = 1 / (x + 3) at x = -3
.3xatcontinuousnotis)3x/(1So3x
1limbut
3x
1lim)c(
.0xatcontinuousnotisxSo
0xfordefinednotisx)b(
.2xatcontinuousis|2x|So
)2(f0|2x|lim|2x|lim)a(
3x3x
2x2x
Example 7 (cont'd)
Example 8Given that f(x) is continuous over the set of all real numbers, find the values of a and b.
2xax2
2x16bx
1xax
)x(f
2
Only need to consider the junctions (x = -1 and x = 2)
1band4a
a46b2So
a4)x(flimand)x(flim6b2)2(f
6ba1So
6b)x(flimand)x(flima1)1(f
2x2x
1x1x
Example 8 (cont'd)
DifferentiabilityGraphical approach: A function f(x) is said to be differentiable at x = a if there is no 'corner' or 'vertical tangency' at that point.
A function must be continuous (but not sufficient) in order that it may be differentiable at that point.
Example 9 (cont'd)
(c)
y
x1 – 1 – 2 – 3 – 4 – 5
2
4
6
8
– 2
– 4
– 6
– 8
Not continuous (not even defined) at x = -2
Derivative of a FunctionA function is differentiable at a point if it is continuous (not broken), smooth (no corner) and not vertical (no vertical tangency) at that point.
Its derivative is given by (First Principle)
h
xfhxfh
)()(lim
0
Differentiability (cont'd)The gradient of PQ is given by
As Q moves closer and closer to P (i.e. as h tends to 0), the limiting value of the gradient of PQ (i.e. the derivative of f(x) at x) becomes the tangent at P.
h
xfhxf )()(
Differentiability (cont'd)The derivative of a function y = f(x) is denoted by
It also represents the rate of change of y with respect to x.
dy dfor y or or f
dx dx
Example 11(a) Find the gradient function of y = 2x2 using first
principle. Find also the gradient at the point (3, 18).
(b)Use the definition (first principle) to find the derivative of ln x and hence find the derivative of ex.
1234dx
dy
x4
)h2x4(limh
h2hx4lim
h
x2)hhx2x(2lim
h
x2)hx(2lim
dx
dy)a(
3x
0h
2
0h
222
0h
22
0h
Example 11 (cont'd)
Example 11 (cont'd)
x
1
elnx
1
)h
xmwhere(])
m
11(lim[ln
x
1
)]x
h1(
h
xlim[ln
x
1)
x
h1(ln
h
1lim
hx
hxln
limh
xln)hx(lnlimxln
dx
d
m
m
0h0h
0h0h
Example 12Find the derivative of the following functions from first principles.
(a) f(x) = 1/x(b) f(x) = x(c) f(x) = xn
Example 12 (cont'd)
1n
1n2n2n
1n
0h
n22n2n
1n
0h
nn
0h
n
nx
)h......hxCnx(lim
h
h......hxChnxlim
h
x)hx(limx
dx
d)c(
ConcavityIf f(x) opens downward, it is said to be concave down
If f(x) opens upwards, it is concave up
concave down concave up
f '(x): + 0 - - 0 +
Point of InflectionPoints of inflection: points where the curve changes from concave up to concave down or concave down to concave up
point of inflection
f '(x): - - - + + + maximum
Horizontal InflectionHorizontal inflection: a point of inflection where the graph is momentarily horizontal, dy/dx = 0
horizontal inflection
-ve
-ve+ve
+ve
Stationary PointsTurning points: max and min
Stationary points: max, min and horizontal inflection
dy/dx = 0
y
x
f '(x)f (x)
a b c
f '(x) changes as below:
x < a: +ve, but
x = a: f '(a) = 0 local max
a < x < b: -ve, , then (less –ve)point of inflection
Example 13 (cont'd)
y
x
f '(x)f (x)
a b c
x = b: f '(b) = 0 local min
b < x < c: +ve, , then point of inflection
x = c: f '(c) = 0 global max
x > c: -ve,
Example 13 (cont'd)
Noteworthy FeaturesMin TP: dy/dx = 0, sign change –ve, 0, +ve
Max TP: dy/dx = 0, sign change +ve, 0, –ve
Horizontal inflection: dy/dx = 0, +ve, 0, +ve or –ve, 0, –ve, (i.e. no sign change)
Point of inflection: d2y/dx2 = 0, (dy/dx is a max/min),
Example 14Graph the following function and its derivative.
Use your graphs to locate the stationary points and points of inflection on
y = x4/4 – 4x3/3 – 7x2/2 + 10x + 5and determine the nature of each.
y = x4/4 – 4x3/3 – 7x2/2 + 10x + 5dy/dx = x3 – 4x2 – 7x + 10
TP dy/dx = 0So x3 – 4x2 – 7x + 10 = 0i.e. (x + 2)(x – 1)(x – 5) = 0 x = -2, 1 or 5
When x = -2, y = -43/3When x = 1, y = 125/12When x = 5, y = -515/12
Example 14 (cont'd)
Piecewise Defined FunctionsA piecewise defined function has different formulas for different parts of its domain.
At junction a filled circle indicates that a point actually exists there, whereas an empty circle shows a discontinuous point.
Example 15Given the function below
(a) Find f(-2), f(1) and f(2)
(b)Graph f and determine whether f is continuous at x = 0 and x = 2.
2xfor1x
2x0forx
0xfor1x/1
)x(f 2
2xfor1x
2x0forx
0xfor1x/1
)x(f 2
(a) f(-2) = 1/(-2) – 1 = -3/2f(1) = 12 = 1f(2) = 2 + 1 = 3
Example 15 (cont'd)
y
x1 2 3 4 – 1 – 2 – 3 – 4
1
2
3
4
5
6
– 1
– 2
– 3
– 4
Example 15 (cont'd)
2xatcontinuousnotis)x(f
)x(flim)x(flim
0xatcontinuousnotis)x(f
)x(flim)x(flim
2x2x
0x0x
Example 16Graph y = | x2 – 6x + 8 | and determine whether the function is continuous at x = 2 and x = 4.
Example 16 (cont'd)
x-1 1 2 3 4 5 6 7
y
-2
2
4
6
8
From graph, the function is continuous at x = 2 and x = 4.
The Sign FunctionIt can be considered as a logical function (especially in computer science)
It extracts the sign of the function
It returns 1 if f(x) is positive, 0 if f(x) equals to 0 and –1 if f(x) is negative.
0)x(f1
0)x(f0
0)x(f1
)x(sgn
(b) y = sgn (x2 – 1)y
x1 2 – 1 – 2
1
2
– 1
– 2
Example 17 (cont'd)
Not continuous at x = -1 and x = 1
Greatest Integer FunctionAlso known as floor functionDefined as the greatest integer less than or equal to the numberThat is, it rounds any number down to the nearest integerSymbol: int [x] orint [4.2] = = 4 int [-2.1] = = -3
x
1.2 2.4
Greatest Integer Function (cont'd)y
x1 2 3 4 – 1 – 2 – 3
1
2
3
– 1
– 2
– 3
Not continuous at all integers.
(a) int [2x –1]
Consider 2x – 1 = n where n is an integerx = (n + 1) / 2So the 'breaking points' are steps of half of an integer
y
x1
2
1 3
2
2 5
2 –
1
2
– 1 –
3
2
– 2
1
2
3
– 1
– 2
– 3
– 4
Example 18 (cont'd)
(b) int [x2]
Consider x2 = n where n is a positive integerx = nSo the breaking points are square root of +ve integers
y
x1
2
1 3
2
2 5
2 –
1
2
– 1 –
3
2
– 2 –
5
2
1
2
3
4 Example 18 (cont'd)
Rules of Differentiation (Review)
2
1nn
gdxdg
fg)dxdf
()
g
f(
dx
d
dx
dgfg)
dx
df()gf(
dx
d
gdx
dfg
dg
d)]x(g[f
dx
d
gdx
df
dx
d)gf(
dx
d
xnxdx
d
Example 19Find the derivative of the following functions:
1xx3
)4x9(y)d(
)3x2(8x
5y)c(
)1x3x(
4y)b(
x
3x7y)a(
3
2
4
22
2
3
3
3
2
4
3142
414
)8x(
)67x6()3x2(5
]8x
3x28[
8x
)3x2(5
8x
)3x2(40
)8x(
)3x2(5
)2()3x2(4)8x(5)3x2()8x)(1(5'y
)3x2()8x(5)3x2(8x
5y)c(
Example 19 (cont'd)
23
223
23
223
3
2
)1xx3(
)1x9()4x9()1xx3)(4x9(18
dx
dy
)1xx3(
)1x9()4x9()1xx3)(9)(4x9(2
dx
dy
1xx3
)4x9(y)d(
Example 19 (cont'd)
x31x2
x31x2
x31x2x3
1x2
x
x
e2
3e4
dx
dy
e)2
3(e)2(2
dx
dy
e2
1e2
e2
1e2y)b(
e3dx
dy
ee3y)a(
Example 20 (cont'd)
]e1
e
ex
x2)
e1
ex(ln2[e
dx
dy
]e1
e
ex
x2[e)]e1(ln)ex([lne2
)]e1(ln)ex([lnee1
exlney)b(
)3x7x
7x2(
e
1
dx
dy
)3x7x(lne
1y)a(
x
x
2x
2x2
x
x
2x2x2x2
x2x2x
2x2
2
2
Example 21 (cont'd)
Differentiability (Revisit)Graphical approach: continuous, no corner, no vertical tangency
Algebraical approach:
)gencytanverticalno(finiteis)a(f
)cornerno()a(f)x(flim)x(flim
)continuous()a(f)x(flim)x(flim
'
''
ax
'
ax
axax
Example 22Determine if the following functions are differentiable at the indicated points.
(a) y = 1 / (x + 1) at x = -1
(b) y = | x + 1 | at x = -1
(c) f(x) = -6x + 5 for x < 3 = -x2 – 4 for x 3 at x = 3
(a) Let f(x) = 1 / (x + 1)
f(-1) is not definedf(x) is not continuous at x = -1f(x) is not differentiable at x = -1
Example 22 (cont'd)
1xatabledifferentinotis)x(f
1)x('flim1)x('flim
1x1
1x1)x('f
1xatcontinuousis)x(f
0)1(f)x(flim)x(flim
1x1x
1x1x)x(f
|1x|)x(fLet)b(
1x1x
1x1x
Example 22 (cont'd)
3xatabledifferentiis)x(f
6)x('flim)x('flim
3xx2
3x6)x('f
3xatcontinuousis)x(f
13)3(f)x(flim)x(flim
3x4x
3x5x6)x(f)c(
3x3x
3x3x
2
Example 22 (cont'd)
Example 23Find the value of a and b so that
f(x) = 3x + 1 for x < 1 = x2 + ax + b for x 1
is continuous and differentiable everywhere.
Possible discontinuity and non-differentiability at x = 1
continuous at x = 1 if 4 = 1 + a + bi.e. a + b = 3
f ’(x) = 3 for x < 1 = 2x + a for x > 1differentiable at x = 1 if 3 = 2 + a
So a = 1 and b = 2
Example 23 (cont'd)
Riemann SumsTo find an approximate area under a curve between two x values [a, b]
The area is divided into n rectangles of equal width
So the width x = (b – a)/n
There are many ways to find the height h of each rectangle (see later)
Then the required area A = hx over the interval [a, b]
Example 24The shaded area below shows the exact area under the curve f(x) = x3 – 3x2 + 8 in the interval [0, 3]
x0.5 1 1.5 2 2.5 3
y
1
2
3
4
5
6
7
8
9
y = x3 – 3x2 + 8
Actual area = 17.25
Example 25(Example 24) Consider n = 5, x = (3 – 0)/5 = 0.6 and h = left endpoint
x = (3 – 0)/5 = 0.6
Left endpoint for the 3rd rectangle
Example 25 (cont'd)f(x1)= (0)3 – 3(0)2 + 8 = 8f(x2)= (0.6)3 – 3(0.6)2 + 8 = 7.136f(x3)= (1.2)3 – 3(1.2)2 + 8 = 5.408f(x4)= (1.8)3 – 3(1.8)2 + 8 = 4.112f(x5)= (2.4)3 – 3(2.4)2 + 8 = 4.544
52.17
)544.4112.4408.5136.78(6.0
x)x(fA5
1ii
Example 26(Example 24) Consider n = 5, x = (3 – 0)/5 = 0.6 and h = right endpoint
Right endpoint for the 3rd rectangle
Example 26 (cont'd)f(x1)= (0.6)3 – 3(0.6)2 + 8 = 7.136f(x2)= (1.2)3 – 3(1.2)2 + 8 = 5.408f(x3)= (1.8)3 – 3(1.8)2 + 8 = 4.112f(x4)= (2.4)3 – 3(2.4)2 + 8 = 4.544f(x5)= (3.0)3 – 3(3.0)2 + 8 = 8
52.17
)8544.4112.4408.5136.7(6.0
x)x(fA5
1ii
Example 27(Example 24) Consider n = 5, x = (3 – 0)/5 = 0.6 and h = minimum point
Minimum point for the 4th rectangle
Example 27 (cont'd)f(x1)= (0.6)3 – 3(0.6)2 + 8 = 7.136f(x2)= (1.2)3 – 3(1.2)2 + 8 = 5.408f(x3)= (1.8)3 – 3(1.8)2 + 8 = 4.112 f(x4)= (2.0)3 – 3(2.0)2 + 8 = 4 f(x5)= (2.4)3 – 3(2.4)2 + 8 = 4.544
12.15
)544.44112.4408.5136.7(6.0
x)x(fA5
1ii
Example 28(Example 24) Consider n = 5, x = (3 – 0)/5 = 0.6 and h = maximum point
Maximum point for the 4th rectangle
Example 28 (cont'd)f(x1)= (0)3 – 3(0)2 + 8 = 8 f(x2)= (0.6)3 – 3(0.6)2 + 8 = 7.136f(x3)= (1.2)3 – 3(1.2)2 + 8 = 5.408f(x4)= (2.4)3 – 3(2.4)2 + 8 = 4.544f(x5)= (3.0)3 – 3(3.0)2 + 8 = 8
8528.19
)8544.4408.5136.78(6.0
x)x(fA5
1ii
Example 29(Example 24) Consider n = 5, x = (3 – 0)/5 = 0.6 and h = midpoint
Midpoint for the 3rd rectangle
Example 29 (cont'd)f(x1)= (0.3)3 – 3(0.3)2 + 8 = 7.757f(x2)= (0.9)3 – 3(0.9)2 + 8 = 6.299f(x3)= (1.5)3 – 3(1.5)2 + 8 = 4.625f(x4)= (2.1)3 – 3(2.1)2 + 8 = 4.031f(x5)= (2.7)3 – 3(2.7)2 + 8 = 5.813
115.17
)813.5031.4625.4299.6757.7(6.0
x)x(fA5
1ii
A Better ApproximationDue to the use of h (left, right, mid, min and max), the rectangles do not truly represent the area under the curve for each strip
If n (number of rectangles) increases, the error decreases
Limit of a SumThe more rectangles, the greater accuracy
So the actual area A is given by
This is written as
x)x(flimAn
egrationintcalledisprocessabovetheand
sumaofitlimtherepresentswhere
dx)x(fx)x(flimAn
Example 32
b 3 4 5 6 7 x
Area
Evaluate the area under the curve y = 3 from x = 2 to x = b by completing the following table.
Hence give an answer for where k is a constant
x
akdx
Example 32 (cont'd)
b 3 4 5 6 7 b
Area 3 6 9 12 15 3(b – 2)
x1 2 3 4 5 6 7 8 9 10
y
1
2
3
4
)ab(kkdxb
a
Evaluate the area under the curve y = 2x from x = 0 to x = b by completing the following table.
Hence give an answer for where k is a constant
Example 33
b 0 1 2 3 4 b
Area
b
0kxdx
Example 33 (cont'd)
b 0 1 2 3 4 x
Area 0 1 4 9 16b2b/2 =
b2
x1 2 3 4 5 6
y
2
4
6
8
10
12
14
2
kbkxdx
2b
0
Given the areas under the curve y = x2 from x = 0 to x = 4 in the following table, find the area when x = b
Hence give an answer for
Example 34
b 0 1 2 3 4 b
Area 0 1/3 8/3 9 64/3
b
0
2dxx