BCC.01.4 – Determining Limits using Limit Laws
Download
Report
Transcript BCC.01.4 – Determining Limits using Limit Laws
A2 – Determining Limits using
Limit Laws and Algebra
IB Math HL&SL - Santowski
(A) Review - The Limit of a Function
The limit concept is the idea that as we get closer
and closer to a given x value in progressively
smaller increments, we get closer to a certain y
value but we never quite reach this y value
We will also incorporate the concept of
"approaching x from both sides" in our discussion
of the concept of limits of a function as we can
approach a given x value either from the right of
the x value or from the left
(A) Review - The Limit of a Function
ex 1. Consider a very simple function of f(x) = x² - 4x + 2 and we will be
asking ourselves about the behaviour of the function near x = 2 (Set up
graphing calculator to see the graph plus tables of values where we make
smaller increments near 2 each time. As we do this exercise, realize that we
can approach the value of x from both the left and the right sides.)
We can present this as lim x2 (x2 – 4x + 2) which we interpret as the fact that
we found values of f(x) very close to -2 which we accomplished by
considering values of x very close to (but not equal to) 2+ (meaning
approaching 2 from the positive (right) side) and 2- (meaning that we can
approach 2 from the negative (left) side)
We will notice that the value of the function at x = 2 is -2 Note that we
could simply have substituted in x = 2 into the original equation to come up
with the function behaviour at this point
(B) Investigating Simple Limit Laws
With our previous example the limit at x = 2 of f(x) = x2 – x + 2 , we
will break this down a bit:
(I) Find the following three separate limits of three separate functions
(for now, let’s simply graph each separate function to find the limit)
lim
lim
lim
(x2) = 4
x2 (-4x) = -4 x lim
x2 (2) = 2
x2
x2
(x) = (-4)(2) = -8
Notice that the sum of the three individual limits was the same as the
limit of the original function
Notice that the limit of the constant function (y = 2) is simply the same
as the constant
Notice that the limit of the function y = -4x was simply –4 times the
limit of the function y = x
(C) Limit Laws
Here is a summary of some important limits laws:
(a) sum/difference rule lim [f(x) + g(x)] = lim f(x) + lim g(x)
(b) product rule lim [f(x) g(x)] = lim f(x) lim g(x)
(c) quotient rule lim [f(x) g(x)] = lim f(x) lim g(x)
(d) constant multiple rule lim [kf(x)] = k lim f(x)
(e) constant rule lim (k) = k
These limits laws are easy to work with, especially when we
have rather straight forward polynomial functions
(D) Limit Laws - Examples
Find lim
x2
(3x3 – 4x2 + 11x –5) using the limit laws
lim x2 (3x3 – 4x2 + 11x –5)
= 3 lim x2 (x3) – 4 lim x2 (x2) + 11 lim x2 (x) - lim x2 (5)
= 3(8) – 4(4) + 11(2) – 5 (using simple substitution or use GDC)
= 25
For the rational function f(x), find
lim x2 (2x2 – x) / (0.5x3 – x2 + 1)
= [2 lim x2 (x2) - lim x2 (x)] / [0.5 lim
= (8 – 2) / (4 – 4 + 1)
=6
x2
(x3) - lim
x2
(x2) + lim
x2
(1)]
(E) Working with More Challenging
Limits – Algebraic Manipulations
But what our rational function from previously was changed slightly
f(x) = (2x2 – x) / (0.5x3 – x2) and we want lim x2 (f(x))
We can try our limits laws (or do a simple direct substitution of x = 2)
we get 6/0 so what does this tell us???
Or we can have the rational function f(x) = (x2 – 2x) / (0.5x3 – x2)
where lim x2 f(x) = 0/0 so what does this tell us?
So, often, the direct substitution method does not work so we need
to be able to algebraically manipulate and simplify expressions to make
the determination of limits easier
(F) Evaluating Limits –
Algebraic Manipulation
Evaluate lim x2 (2x2 – 5x + 2) / (x3 – 2x2 – x + 2)
With direct substitution we get 0/0 ????
Here we will factor first (Recall factoring techniques)
= lim x2 (2x – 1)(x – 2) / (x2 – 1)(x – 2)
= lim x2 (2x – 1) / (x2 – 1) cancel (x – 2) ‘s
Now use limit laws or direct substitution of x = 2
= (2(2) – 1) / ((2)2 – 1))
= 3/3
=1
(F) Evaluating Limits –
Algebraic Manipulation
1 1
Evaluate
lim x 3
x 3 x 3
Strategy was to find
a common denominator
with the fractions
1
1
3
lim x
x 3 x 3
3
x
3x
lim 3x
x 3
x 3
3 x
3x
lim
x 3 1( 3 x )
3 x
lim
x 3 3x ( 3 x )
1
lim
19
x 3 3x
(F) Evaluating Limits –
Algebraic Manipulation
x 4
Evaluate lim
x 4
x2
(we recall our earlier
work with complex numbers
and conjugates as a way
of making “terms disappear”
x 4
lim
x 4
x2
lim
x 2
x 4 x 2
lim
x 4
x 4
x 4 x 2
x2
( x 4)
x 2
lim
x 4
1
6
(G) Internet Links
Limit Properties - from Paul Dawkins at
Lamar University
Computing Limits - from Paul Dawkins
at Lamar University
Limits Theorems from Visual Calculus
Exercises in Calculating Limits with
solutions from UC Davis
(H) Homework
Stewart, 1989, Calculus – A First
Course, Chap 1.2, p19, Q3-6eol, 7,8,9