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 x  2 (x 2 – 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) = x 2 – 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 (x 2 ) = 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 (3x 3 – 4x x –5) using the limit laws lim x  2 (3x 3 – 4x x –5) = 3 lim x  2 (x 3 ) – 4 lim x  2 (x 2 ) + 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 (2x 2 – x) / (0.5x 3 – x 2 + 1) = [2 lim x  2 (x 2 ) - lim x  2 (x)] / [0.5 lim x  2 (x 3 ) - lim x  2 (x 2 ) + 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) = (2x 2 – x) / (0.5x 3 – x 2 ) 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) = (x 2 – 2x) / (0.5x 3 – x 2 ) 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 (2x 2 – 5x + 2) / (x 3 – 2x 2 – x + 2) With direct substitution we get 0/0  ???? Here we will factor first (Recall factoring techniques) = lim x  2 (2x – 1)(x – 2) / (x 2 – 1)(x – 2) = lim x  2 (2x – 1) / (x 2 – 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 Handouts from other textbooks Calculus, a First Course, J. Stewart, p19, Q1-6 eol