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A2 – Determining Limits using Limit Laws and Algebra

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A2 – Determining Limits using Limit Laws and Algebra. IB Math HL&amp;SL - Santowski. (A) Review - The Limit of a Function.

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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
• Evaluate
• Strategy was to find

a common denominator

with the fractions

(F) Evaluating Limits – Algebraic Manipulation
• Evaluate

(we recall our earlier

work with complex numbers

and conjugates as a way

of making “terms disappear”

(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