# A2 � Determining Limits using Limit Laws and Algebra

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```					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 x2 (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   x2(x2) = 4
lim   x2 (-4x) = -4 x lim   x2   (x) = (-4)(2) = -8
lim   x2 (2) = 2

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   x2   (3x3 – 4x2 + 11x –5) using the limit laws

lim x2 (3x3 – 4x2 + 11x –5)
= 3 lim x2 (x3) – 4 lim x2 (x2) + 11 lim x2 (x) - lim x2 (5)
= 3(8) – 4(4) + 11(2) – 5 (using simple substitution or use GDC)
= 25

For the rational function f(x), find
lim x2 (2x2 – x) / (0.5x3 – x2 + 1)
= [2 lim x2 (x2) - lim x2 (x)] / [0.5 lim   x2   (x3) - lim   x2   (x2) + lim   x2   (1)]
= (8 – 2) / (4 – 4 + 1)
=6
(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 x2 (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 x2 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 x2 (2x2 – 5x + 2) / (x3 – 2x2 – x + 2)
With direct substitution we get 0/0  ????

Here we will factor first (Recall factoring techniques)
= lim x2 (2x – 1)(x – 2) / (x2 – 1)(x – 2)
= lim x2 (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                    1

1
lim x 3        lim x
x 3 x  3
3
x 3 x  3
3       x

Strategy was to find    lim 3x        3x
x 3    x 3
a common denominator                 3 x
with the fractions         lim       3x
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                 x 4
Evaluate lim                 lim
x 4 x2             x 4 x2
 x  4   x  2
 lim
(we recall our earlier
work with complex numbers
x 4           x  2
x2 
and conjugates as a way              x  4   x  2
of making “terms disappear”    lim
x 4           ( x  4)
x 2
 lim                  6
x 4  1
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

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