Unformatted text preview: MTH405 Wk 7 Lect 1 Calculus
Background: Limits of Functions
Denition. (Limit of a function) Let f be a function on the real numbers and c be an element in R. Then limit of f as x approaches c is dened to be the
value f (x) as x comes very close to c. We denote this by
lim f (x). x→c Denition. (Directional Limit) Let f be a function on the real numbers and c be an element in R. Then limit of f as x approaches c from the positive side
is dened to be the value f (x) as x comes very close to c from the right end of
the graph (moving from right to left and approaching c). We denote this by
lim f (x). x→c+ Denition. (Directional Limit) Let f be a function on the real numbers and c
be an element in R. Then limit of f as x approaches c from the negative side is
dened to be the value f (x) as x comes very close to c from the left end of the
graph (moving from left to right and approaching c). We denote this by
lim f (x). x→c− Theorem.
Then Let f be a function on the
limx→c f (x) exists if and only if real numbers and lim f (x) = k = lim− f (x). x→c+ x→c Hence, lim f (x) = k. x→c
However, if lim f (x) 6= lim f (x),
x→c− x→c+
then lim f (x) x→c does not exist. 1 c be an element in R. Remark. There are 3 important points to note: • The limit is basically the y value for some x value in a function.
• The limit of a function f (x) as x → c is to nd the value where the y
value should be as x goes to c. Remember, we are not trying to nd the
actual value of function f (x) at x = c, but what it should be, by looking at the behaviour of the nearby points. • Most of the time we will have the value of f (x) evaluated at c to be exactly
limx→c f (x). But there are cases where the value of f (x) itself may not
be dened at c, but even then limx→c f (x) exists. (This is because of the point above) Types of Approaches of x → c
There are 5 dierent types of approaches.
• limx→c f (x).
• limx→c+ f (x).
• limx→c− f (x).
• limx→+∞ f (x).
• limx→−∞ f (x). Types of Solutions for Limits
There are 3 types of solutions that you can have when evaluating limits.
• A value in real numbers (L)
• Innity (either +∞ or −∞)
• Limit does not exist. 2 Properties of Limits
Theorem. Let c and k be real numbers and assume that limx→c f (x) = L1
holds for functions f (x) and g(x). limx→c g(x) = L2 Then 1. limx→c k = k.
2. limx→c [f (x) + g(x)] = L1 + L2 .
3. limx→c [f (x) − g(x)] = L1 − L2 .
4. limx→c f (x)g(x) = L1 · L2 .
5. limx→c [k · f (x)] = k = kL1 .
(x)
=
6. limx→c fg(x) L1
L2 , provided L2 6= 0. 7. limx→c f (x)n = [limx→c f (x)]n = Ln .
8. limx→c p
n f (x) = p
n limx→c f (x) = 3 √
n L1 . Finding Limits
There are 3 methods to nd limits that we will learn.
• Graphical method
• Tabular method
• Algebraic method (or by calculation) Graphical Method
In this method, we draw graphs to nd the limit of f (x) as x → c. Example. Evaluate the limits of the functions below using graphical method.
1. limx→2 3x + 2.
2. limx→3 3x + 2.
3. limx→0 4x2 . Tabular method
This is the easiest of the three methods when evaluating limits and it can be
used separately to verify the answers when using your nding limits using other
methods.
In this method, we construct tables with 2 rows: one for x values and one
for the corresponding y values, as x → c.
• For the x values, we input the values coming closer and closer to c.
• Then using the x values, we calculate y , except when x = c which we leave out blank. • Then looking at the behaviour of all the nearby points, we make a decision
about the limit of f (x) as x → c. Example. Evaluate the limits of the following functions using the tabular
method. 1. limx→5 3x + 2.
2. limx→0 sinx x . 4 Algebraic Method
When evaluating
lim f (x), x→c the general procedure is nding limits is to
1. First substitute c inside f (x).
2. If answer is of the form a
b and 0
b = 0, then that is the answer. 3. If the answer is a0 = , then we have to check if limit is +∞, or −∞ or
if the limit does not exist by using a Sign Test.
4. If the answer is of the form 00 (this form actually means f (x) can be
factorized and simplied), then go on to factorize and simplify the function
f (x). And then substitute c again. Example. Find the following limits.
1. limx→1 3x + 5.
2. limx→3 2x2 .
−4
3. limx→2 xx−2
.
2 −9
.
4. limx→3 xx−3
2 5. limx→3 x 2 −2x−3
x−3 5 Calculus (Dierentiation)
Basic Question: What is Calculus? Calculus is the study of how fast or how slow things change. In
other words, Calculus is the study of the rate of change.
Many ideas in Calculus originated with the following two geometric
problems: THE TANGENT LINE PROBLEM. Given a function f and a point P = (x0 , y0 ) on its graph, nd an equation of the line that is
tangent to the graph at point P .
Solving the tangent line problem, we end up with
Dierentiation. techniques of THE AREA PROBLEM. Given a function f , nd the area between the graph of f and an interval [a, b] on the xaxis.
Solving the Area problem, we end up with the
tegration. 6 techniques of In The Tangent Line Problem
Given a function f and a point P = (x0 , y0 ) on its graph,
nd an equation of the line that is tangent to the graph at
point P . So we have two points P = (x0 , y0 ) and Q = (x1 , y1 ) on the
function f .
• The only way to nd the gradient of the tangent line touching
at P is to gradually bring the line touching P and Q (blue)
closer to tangent line (pink).
• So we start with the gradient of the blue line:
mblue = • f (x1 ) − f (x0 )
.
x1 − x0 As the blue line comes closer to the pink, we see that point Q
approaches point P . In other words,
mblue = lim x1 →x0 • f (x1 ) − f (x0 )
.
x1 − x0 Now, we see that x1 → x0 , that means h = x1 − x0 approaches
0. Hence, we can improve the function
f (x1 ) − f (x0 )
x1 →x0
x1 − x0
f (x1 ) − f (x0 )
=
lim
(x1 −x0 )→0
x1 − x0
f (x0 + h) − f (x0 )
= lim
.
h→0
h mblue = • lim And from this began Calculus. (Both Newton and Leibzig
made this approach independently)
7 Denition of the derivative
Denition. Let y = f (x) be a function on the real numbers and f 0
is the derivative of f . Then we use the denition of the derivative
to nd the derivative of f . The denition of the derivative formula
is given by
dy
f (x + h) − f (x)
= f 0 (x) = lim
.
h→0
dx
h
Remark. The procedure to use Denition of the Derivative is: • to rst nd and expand f (x) and f (x + h) • then substitute into the denition formula and expand and simplify the numerator. • After that, we nd h common in numerator and factorize it
outside. Then cancel it with the h in the denominator. • Once h in the denominator is cancelled, then nd the limit as
h → 0 (this should always be the last thing to do) to nd the
answer. Example. Use the Denition of the Derivatives to nd the derivative
of the following functions.
1. f (x) = 2x.
2. f (x) = 3x + 5.
3. f (x) = 2x2
4. f (x) = 3x2 + x 8 Solution.
1. We have f (x) = 2x and f (x + h) = 2(x + h) = 2x + 2h. Then (2x + 2h) − (2x)
h→0
h
2h
= lim
h→0 h
= lim 2 = 2. f 0 (x) = lim h→0 2. We have f (x) = 3x+5 and f (x+h) = 3(x+h)+5 = 3x+3h+5. Then (3x + 3h + 5) − (3x + 5)
h
3h
= lim
h→0 h
= lim 3 = 3. f 0 (x) = lim h→0 h→0 3. We have f (x) = 2x2 and f (x + h) = 2(x + h)2
= 2(x2 + 2xh + h2 )
= 2x2 + 4xh + 2h2 . Then
(2x2 + 4xh + 2h2 ) − (2x2 )
h→0
h
2
4xh + 2h
= lim
h→0
h
h(4x + 2h)
= lim
h→
h
lim (4x + 2h) = 4x. f 0 (x) = lim h→0 9 4. We have f (x) = 3x2 + x and f (x + h) = 3(x + h)2 + (x + h)
= 3(x2 + 2xh + h2 ) + x + h
= 3x2 + 6xh + 3h2 + x + h. Then
(3x2 + 6xh + 3h2 + x + h) − (3x2 + x)
h→0
h
3h2 + 6xh + h
= lim
h→0
h
h(3h + 6x + 1)
= lim
h→0
h
= lim 3h + 6x + 1 = 6x + 1. f 0 (x) = lim h→0 10...
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 Spring '16
 Alveen Chand
 Calculus, Real Numbers, Derivative, lim, Continuous function

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