1 / 12

BCC.01.4 – Determining Limits using Limit Laws and Algebra

BCC.01.4 – Determining Limits using Limit Laws and Algebra. MCB4U - Santowski. (A) Review - The Limit of a Function.

nasia
Download Presentation

BCC.01.4 – Determining Limits using Limit Laws and Algebra

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. BCC.01.4 – Determining Limits using Limit Laws and Algebra MCB4U - Santowski

  2. (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

  3. (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

  4. (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

  5. (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

  6. (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

  7. (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

  8. (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

  9. (F) Evaluating Limits – Algebraic Manipulation • Evaluate • Strategy was to find a common denominator with the fractions

  10. (F) Evaluating Limits – Algebraic Manipulation • Evaluate (we recall our earlier work with complex numbers and conjugates as a way of making “terms disappear”

  11. (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

  12. (H) Homework • Handouts from other textbooks • Calculus, a First Course, J. Stewart, p19, Q1-6 eol

More Related