LIMITS AND CONTINUITY

Calculus Vol. 1 CONTENTS CHAPTER 1: LIMITS AND CONTINUITY Pages 2 – 16 1.1. THE LIMIT OF A FUNCTION 2 1.2. THE LIMIT LAWS 7 1.3. CONTINUITY 13 CHAPTER 2: DERIVATIVES Pages 17 – 52 2.1 DEFINING THE DERIVATIVE 17 2.2 THE DERIVATIVE AS A FUNCTION 22 2.3 DIFFERENTIATION RULES 26 2.4 DERIVATIVES AS A RATE OF CHANGE 39 2.5 THE CHAIN RULE 42 2.6 IMPLICIT DIFFERENTIATION 46 2.7 PARAMETRIC DIFFERENTIATION CHAPTER 3: APPLICATIONS OF DERIVATIVES Pages 53 – 80 3.1 RELATED RATES 53 3.2 MAXIMA AND MINIMA 60 3.3 THE MEAN VALUE THEOREM 66 3.4 DERIVATIVES AND THE SHAPE OF A GRAPH 71 CHAPTER 4: ANTIDERIVATIVES AND INTEGRATION Pages 81 – 104 4.1 ANTIDERIVATIVES 81 4.2 THE DEFINITE INTEGRAL 92 4.3 INTEGRATION BY SUBSTITUTION 96 4.4 INTEGRATION BY PARTS 100 Calculus Vol. 1 2 CHAPTER 1 LIMITS AND CONTINUITY INTRODUCTION The idea of a limit is central to all of calculus. In this chapter, we describe how to find the limit of a function at a given point. Not all functions have limits at all points, and we discuss what this means and how we can tell if a function does or does not have a limit at a particular value. This chapter has been created in an informal, intuitive fashion, but this is not always enough if we need to prove a mathematical statement involving limits. 1.1 THE LIMIT OF A FUNCTION The concept of a limit or limiting process, essential to the understanding of calculus, has been around for thousands of years. In fact, early mathematicians used a limiting process to obtain better and better approximations of areas of circles. Yet, the formal definition of a limit – as we know and understand it today – did not appear until the late 19th century. We therefore begin our quest to understand limits, as our mathematical ancestors did, by using an intuitive approach. We begin our exploration of limits by looking at the graph of the x2  4 given in Figure 1.1. x 2 This function is undefined at x = 2, but if we make this statement and no other, we give a very incomplete picture of how each function behaves in the vicinity of x = 2. To express the behavior the graph in the vicinity of 2 more completely, we need to introduce the concept of a limit. function f ( x)  Let’s first take a closer look at how the function behaves around x  2 in Figure 1.1. As the values of x approach from either side of 2, the values of y  f ( x) approach 4. Mathematically, we say that the limit of f  x  as x approaches 2 is 4. Symbolically, we express this limit as lim f  x   4 . x 2 Definition Let f  x  be a function defined at all values in an open interval containing a , with the possible exception of a itself, and let L be a real number. If all values of the function f  x  approach the real number L as the values of x (  a ) approach the number a, then we say that the limit of f  x  as x approaches a is L (More succinct, as x gets closer to a , f  x  gets closer and stays close to L ). Symbolically, we express this idea as lim f  x   L . x a Download for free at https://openstax.org/details/books/calculus-volume-1. Calculus Vol. 1 3 Example 1.1: Evaluating a Limit Using a Table of Functional Values Evaluate lim x4 x 2 using a table of functional values. x4 Solution We use Table 1.1 to list the values of the function for the given values of x. x 3.9 3.99 3.999 3.9999 3.99999 x 2 x4 0.251582342 0.250156446 0.250015627 0.250001563 0.25000016 f  x  x 2 x4 0.248456731 0.249843945 0.249984377 0.249998438 0.24999984 f  x  x 4.1 4.01 4.001 4.0001 4.00001 x 2 x4 x  4 After inspecting this table, we see that the functional values less than 4 appear to be decreasing toward 0.25 whereas the functional values greater than 4 appear to be increasing toward 0.25. We Table 1.1: Table of functional values for lim x 2  0.25 . x 4 x  4 Note that for the limit of a function to exist at a point, the functional values must approach a single real-number value at that point. If the functional values do not approach a single value, then the limit does not exist. conclude that lim Example 1.2: Evaluating a Limit that fail to exist x 3 Evaluate lim using a table of functional values. x 3 x  3 Solution We use Table 1.2 to list the values of the function for the given values of x. x 2.9 2.99 2.999 2.9999 2.99999 f  x  -1 -1 -1 -1 -1 x 3 x 3 x f  x  3.1 3.01 3.001 3.0001 3.00001 1 1 1 1 1 x 3 x 3 x 3 x 3 x  3 After inspecting this table, we see that the functional values less than 3 is equal to  1 , whereas the x 3 functional values greater than 3 is equal to 1 . We conclude that lim does not exist. x 3 x  3 Table 1.2: Table of functional values for lim Download for free at https://openstax.org/details/books/calculus-volume-1. Calculus Vol. 1 4 Example 1.3: Evaluating a Limit Using a Graph For g  x  shown in Figure 1.2, evaluate lim g  x  . x  1 Solution: Despite the fact that g  1  4 , as the x -values approach −1 from either side, the g  x  values approach 3. Therefore, lim g  x   3 . Note that we can determine this x  1 limit without even knowing the algebraic expression of the function. Based on Example 1.3, we make the following observation: It is possible for the limit of a function to exist at a point, and for the function to be defined at this point, but the limit of the function and the value of the function at the point may be different. One-Sided Limits Sometimes indicating that the limit of a function fails to exist at a point does not provide us with enough information about the behavior of the function at that particular point. To see this, we now look at the x2 function g  x   given in Figure 1.3. As we x2 pick values of x close to 2, g  x  does not approach a single value, so the limit as x approaches 2 does not exist – that is, lim g  x  DNE. However, this x 2 statement alone does not give us a complete picture of the behavior of the function around the x -value 2. To provide a more accurate description, we introduce the idea of a one-sided limit. For all values to the left of 2 (or the negative side of 2), g  x  = −1. Thus, as x approaches 2 from the left, g  x  approaches −1. Mathematically, we say that the limit as x approaches 2 from the left is −1. Symbolically, we express this idea as lim g  x   1 . x2  Similarly, as x approaches 2 from the right (or from the positive side), g  x  approaches 1. Symbolically, we express this idea as lim g  x   1 x2  We can now present an informal definition of one-sided limits. Download for free at https://openstax.org/details/books/calculus-volume-1. Calculus Vol. 1 5 Definition: We define two types of one-sided limits. Limit from the left: Let f  x  be a function defined at all values in an open interval of the form ( c , a ), and let L be a real number. If the values of the function f  x  approach the real number L as the values of x (where x  a ) approach the number a , then we say that L is the limit of f  x  as x approaches a from the left. Symbolically, we express this idea as lim f  x   L . xa Limit from the right: Let f  x  be a function defined at all values in an open interval of the form ( a , c ), and let L be a real number. If the values of the function f  x  approach the real number L as the values of x (where x  a ) approach the number a , then we say that L is the limit of f  x  as x approaches a from the right. Symbolically, we express this idea as lim f  x   L . xa Theorem 1.1: Relating One-Sided and Two-Sided Limits Let f  x  be a function defined at all values in an open interval containing a , with the possible exception of a itself, and let L be a real number. Then, lim f  x   L if and only if x a lim f  x   L and lim f  x   L . xa  xa Example 1.4: Evaluating One-Sided Limits  x  1 if For the function f  x    2  x  4 if of functional values. (a) lim f  x  x2 x2 , evaluate each of the following limits using a table x2 (b) lim f  x  x2 Solution We can use tables of functional values again Table 1.3. Observe that for values of x less than 2, we use f  x   x  1 and for values of x greater than 2, we use f  x   x2  4 . x 1.9 1.99 1.999 1.9999 1.99999 f  x  x 1 f  x   x2  4 x 2.1 0.41 2.01 0.0401 2.001 0.004001 2.0001 0.00040001 2.00001 0.0000400001 x2  x  1 if Table 1.3: Table of functional values for f  x    2 x2  x  4 if Based on this table, we can conclude that (a) lim f  x  = 3 and x2 2.9 2.99 2.999 2.9999 2.99999 (b) lim f  x  = 0. x2 Therefore, the (two-sided) limit of f  x  does not exist at x = 2. Download for free at https://openstax.org/details/books/calculus-volume-1. Calculus Vol. 1 6 EXERCISE 1.1 1 1 x 1. Estimate lim using a table of functional values. x 1 x  1 2. Use the graph of h(x) in Figure 1.1.1 to evaluate lim h  x  , if possible. x  2 3. Use a table of functional values to evaluate lim x 2 x2  4 x2 , if possible. 4. Use a table of functional values to evaluate lim x0 sin x , if x possible. 5. Use a table of functional values to estimate the following limits, if possible. (a) lim x 1 (b) lim x2 1 x 1 x2 1 x 1 6. Use the graph of f  x  in Figure 1.1.2 to determine each x 1 of the following values: (a) lim  f  x  ; lim  f  x  ; lim f  x  ; f  4 x  4 x  4 x  4 (b) lim  f  x  ; lim  f  x  ; lim f  x  ; f  2 x  2 x  2 x  2 (c) lim f  x  ; lim f  x  ; lim f  x  ; f 1 x 1 x 1 x 1 (d) lim f  x  ; lim f  x  ; lim f  x  ; f  3 x 3 x 3 x 3 7. In the following exercises, consider the graph of the function y  f  x  shown in Figure 1.1.3. Which of the statements about y  f  x  are true and which are false? Explain why a statement is false. (a) lim f  x   0 x  10 (b) lim f  x   3 x 2 (c) lim f  x   f  8  x  8 (d) lim f  x   5 x 6 8. In the following exercises, use the graph of the function y  f  x  shown in Figure 1.1.4 to find the values, if possible. Estimate when necessary. Download for free at https://openstax.org/details/books/calculus-volume-1. Calculus Vol. 1 7 (a) lim  f  x  x 2 (b) lim  f  x  x 2 (c) lim f  x  x 2 (d) lim f  x  x 2 (e) lim f  x  x 2 (f) lim f  x  x 2 9. In the following exercises, use the graph of the function y  f  x  shown in Figure 1.1.5 to find the values, if possible. Estimate when necessary. (a) lim f  x  x 0 (b) lim f  x  x 0 (c) lim f  x  x 0 (d) lim f  x  x 1 (e) lim f  x  x 2 10. Using the graph in Figure 1.1.6, find each limit or explain why the limit does not exist. (a) lim f  x  x  1 (b) lim f  x  x 1 (c) lim f  x  x0 (d) lim f  x  x2 1.2 THE LIMIT LAWS In the previous section, we evaluated limits by looking at graphs or by constructing a table of values. In this section, we establish laws for calculating limits and learn how to apply these laws. We begin by restating two useful limit results from the previous section. These two results, together with the limit laws, serve as a foundation for calculating many limits. Theorem 1.2: Basic Limit Results For any real number a and any constant c , (i) lim x  a x a (ii) lim c  c x a Download for free at https://openstax.org/details/books/calculus-volume-1. Calculus Vol. 1 8 Theorem 1.3: Limit Laws Let f  x  and g  x  be defined for all x  a over some open interval containing a . Assume that L and M are real numbers such that lim f  x   L and lim g  x   M . Let c be a constant. x a x a Then, each of the following statements holds: Sum law for limits: lim  f  x   g  x    lim f  x   lim g  x   L  M . x a x a x a Difference law for limits: lim  f  x   g  x    lim f  x   lim g  x   L  M . x a x a x a Constant multiple law for limits: lim  c f  x    c  lim f  x   c  L . x a x a Product law for limits: lim  f  x   g  x    lim f  x   lim g  x   L  M . x a x a x a f  x L  f  x   xlim a   Quotient law for limits: lim  for M  0 .   x a g  x M  g  x   xlim a  Power law for limits: lim  f  x    lim f  x  n x a Root law for limits: lim x a n x a   L for every positive integer n . n n f  x   n lim f  x   n L for all L if n is odd and for L  0 if n is x a even. Theorem 1.4: Limits of Polynomial and Rational Functions Let p  x  and q  x  be polynomial functions. Let a be a real number. Then, lim p  x   p  a  x a p  x p a  when q  a   0 . x a q  x  q a lim Example 1.5: Evaluating Limits using the limit laws 2 x 2  3x  1 . x 2 x3  4 Use the limit laws to evaluate lim Solution 2 x 2  3x  1 2  2   3  2   1 3 lim   . 3 x 2 x3  3 11  2  3 2 Additional Limit Evaluation Techniques Calculating a Limit when f  x 0 has the Indeterminate Form . 0 g  x 1. First, we need to make sure that our function has the appropriate form and cannot be evaluated immediately using the limit laws.` f  x 2. We then need to find a function that is equal to h  x   for all x  a over some interval g  x containing a . To do this, we may need to try one or more of the following steps: Download for free at https://openstax.org/details/books/calculus-volume-1. Calculus Vol. 1 9 If f  x  and g  x  are polynomials, we should factor each function and cancel out any  common factors. If the numerator or denominator contains a difference involving a square root, we should try multiplying the numerator and denominator by the conjugate of the expression involving the square root. f  x If is a complex fraction, we begin by simplifying it. g  x   3. Last, we apply the limit laws. Example 1.6: Evaluating a Limit by Factoring and Canceling x 2  3x . x  3 2 x2  5x  3 Evaluate lim Solution Step 1. The function f  x   function we get x 2  3x is undefined for x  3 . In fact, if we substitute 3 into the 2 x2  5x  3 0 , which is undefined. Factoring and canceling is a good strategy: 0 x  x  3 x 2  3x lim 2  lim x  3 2 x  5x  3 x  3  x  3 2 x  1 Step 2. For all x  3 , x 2  3x x . Therefore,  2 2 x  5x  3 2 x  1 x  x  3 x  lim x  3  x  3 2 x  1 x  3  2 x  1 lim Step 3. Evaluate using the limit laws: x 3  . x  3  2 x  1 7 lim Example 1.7: Evaluating a Limit by Multiplying by a Conjugate Evaluate lim x  1 x  2 1 . x 1 Solution 0 x  2 1 has the form at  1 . Let’s begin by multiplying by 0 x 1 x  2  1 , on the numerator and denominator: Step 1. of lim x  1 x  2  1 , the conjugate x  2 1 x  2 1 x  2 1  lim  . x  1 x 1 x 1 x  2 1 Step 2. We then multiply out the numerator. We don’t multiply out the denominator because we are hoping that the ( x  1 ) in the denominator cancels out in the end: x 1 . lim x  1  x  1 x  2  1   Download for free at https://openstax.org/details/books/calculus-volume-1. Calculus Vol. 1 10 Step 3. Then we cancel: lim x  1  1  .  1 . 2 x  2 1 Step 4. Last, we apply the limit laws: lim x  1  1  x  2 1 Example 1.8: Evaluating a Limit by Simplifying a Complex Fraction 1 1  Evaluate lim x  1 2 x 1 x 1 Solution 1 1  0 Step 1. x  1 2 has the form at 1. We simplify the algebraic fraction 0 x 1 2   x  1 1 1  2  x  1   x  1 x 1 . lim x  1 2  lim  lim  lim x 1 x 1 x 1 2  x  1 x  1 x 1 2  x  1 x  1 x 1 x 1 Step 2. Then, we cancel the common factors of  x  1 : 1 . x 1 2  x  1 Step 3. Last, we evaluate using the limit laws: 1 1 lim  . x 1 2  x  1 4  lim Example 1.9: Evaluating a Two-Sided Limit Using the Limit Laws  4 x  3 if x  2 For f  x    , evaluate each of the following limits: 2   x  3 if x  2 a. lim f  x  b. x2  c. lim f  x  lim f  x  x2  x2 Solution a. Since f  x   4x  3 for all x  2 , replace f  x  in the limit with 4 x  3 and apply the limit laws: lim f  x   lim  4 x  3  5 . x2 x2 b. Since f  x    x  3 for all x  2 , replace f  x  in the limit with  x  3 and apply the limit laws: 2 lim f  x   lim  x  3  1 . 2 2 x 2 x 2 c. Since lim f  x   5 and lim f  x   1, we conclude that lim f  x  does not exist. x2 x 2 x2 Download for free at https://openstax.org/details/books/calculus-volume-1. Calculus Vol. 1 11 We now turn our attention to evaluating a limit of the form lim xa where K  0 and lim g  x   0 . That is xa f  x K , has the form , K  0 at a . 0 g  x Example 1.10: Evaluating a Limit of the Form Evaluate lim x2 f  x , where lim f  x   K , xa g  x K , K  0 using the Limit Laws 0 x3 . x  2x 2 Solution 1 . That is, as x 0 approaches 2 from the left, the numerator approaches −1; and the denominator approaches 0. Step 1. After substituting in x  2 , we see that this limit has the form x 3 becomes infinite. To get a better idea of what the limit x  x  2 is, we need to factor the denominator: x 3 x 3 lim 2  lim . x2 x  2x x2 x  x  2 Step 2. Since x  2 is the only part of the denominator that is zero when 2 is substituted, we then 1 separate from the rest of the function: x2 x3 1  lim  x2 x x2 x  3 1 1 x3 1    . Therefore, the product of  Step 3. lim and lim and has a x2 x  2 x2 x2 x x 2 limit of   : x3 lim 2   . x2 x  2x Consequently, the magnitude of    EXERCISE 1.2 1. In the following exercises, use direct substitution to evaluate each limit. (a) lim  4 x 2  1 x2 2 x 2  3x  1 x 3 5x  4 1 (c) lim x  0 1  sin x 2  7x (d) lim x 1 x  6 (b) lim (e) lim e2 x x 2 2. In the following exercises, use direct substitution to show that each limit leads 0 to the indeterminate form . Then, 0 evaluate the limit. x 2  16 (a) lim x4 x  4 x2 (b) lim 2 x2 x  2x 3 x  18 (c) lim x  6 2 x  12 x 2 (f) lim ln e x 3 3x (d) lim h 0 1  h  2 1 h Download for free at https://openstax.org/details/books/…
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