BCC.01.4 – Determining Limits using Limit Laws

and Algebra

MCB4U - 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 x2 (x2) = 4

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

lim x2 (2) = 2

Notice that the sum of the three individual limits was the same as the limit of the original functionNotice that the limit of the constant function (y = 2) is simply the same as the constantNotice that the limit of the function y = -4x was simply –4 times the limit of the function y = x

(C) Limit LawsHere 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 - ExamplesFind 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 x2 (x3) - lim x2 (x2) + lim x2 (1)]= (8 – 2) / (4 – 4 + 1)= 6

(E) Working with More Challenging Limits – Algebraic

ManipulationsBut 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

Evaluate

Strategy was to find a common denominatorwith the fractions

limx

xx

3

1 133

lim

lim

lim( )

lim( )

lim

x

x

x

x

x

xx

xxx

xxxx

x

x x

x

3

3

3

3

3

1 133

33 3

33

31 3

3

3 3

1

31

9

(F) Evaluating Limits – Algebraic Manipulation

Evaluate

(we recall our earlierwork with complex numbersand conjugates as a wayof making “terms disappear”

limx

x

x

4

4

2

lim

lim

lim( )

lim

x

x

x

x

x

x

x x

x x

x x

x

x

4

4

4

4

4

2

4 2

2 2

4 2

4

2

16

(G) Internet LinksLimit Properties - from Paul Dawkins at Lamar University Computing Limits - from Paul Dawkins at Lamar UniversityLimits Theorems from Visual CalculusExercises in Calculating Limits with solutions from UC Davis

(H) HomeworkHandouts from other textbooks

Calculus, a First Course, J. Stewart, p19, Q1-6 eol