Introduction to Limits

Transcription

1 MATH 136 Introduction to Limits Given a function y = f (x), we wish to describe the behavior of the function as the variable x approaches a particular value a. We should be as specific as possible in describing the behavior of both x and the function values f (x). For instance, x may be increasing to a (i.e., approaching a from the left ) or x may be decreasing to a (i.e., approaching a from the right + ). Likewise, the function values may be either increasing or decreasing. Example 1. Evaluate and explain the following its: (a) 4sin x (b) + 4sin x (c) 4 sin x Solution. We can evaluate the its by simply computing 4 sin(7π /6) and then describe the behavior by observing the graph. 4 sin x = 4 sin(7π /6) = 4 (1 / 2) = 2 As x increases to 7π /6, then 4sin x decreases to 2. Likewise, 4sin x = 2 x 7π +. But now, as x decreases to 7π /6, then 4sin x increases to 2. 6 Because these one-sided its are equal to the same value and are finite, we say that 4 sin x exists and 4 sin x = 2 also. For this last it we say, as x approaches 7π /6, then 4sin x approaches 2. If the graph is continuous (i.e., no holes, jumps, asymptotes, etc.), then the y values will approach the function value f (a) as x approaches a (i.e., f ( x) = f (a) ). We just need to specify how the y values increase/decrease to f (a) as x is increasing to a (from the left) and as x is decreasing to a (from the right). It is often helpful to graph the function on your calculator and then use one of several built-in commands to compute the function value. Example 2. Let f (x) = x x 2 4. Evaluate and describe the its: (a) f (x) (b) f (x) + (c) f (x)

2 Solution. Because f (x) = x x 2 4 is a continuous polynomial, we can evaluate each it as x approaches 2 by simply evaluating f (2). Then we can observe the graph to describe the behavior. After graphing with an appropriate WINDOW, it is easy to observe the behavior of the y values. Again because f is continuous, the y values will approach the function value f (2) as x approaches 2. We now explain two ways to compute this function value on your calculator: On TI-4: After graphing the function in Y1, press 2nd TRACE (i.e. CALC). Press 1 for value. Enter 2 for X. Then f (2) is given as Y =. On TI-9: After graphing the function in y1 (in APPS, Y= Editor), press F5 for MATH, then press 1 for Value. Enter 2 for xc. Then f (2) is given as yc =. Now: f (x) = As x increases to 2, then x x 2 4 increases to. + f (x) = As x decreases to 2, then x x 2 4 decreases to. f (x) = As x approaches to 2, then x x2 4 approaches. Second Method of Function Evaluation On TI-4: After entering the function in Y1, press 2nd Quit to return to the Home screen. Then press VARS. Scroll right to Y-VARS. Press 1 for Function. Press 1 for Y1. Then Y1 comes up on the Home screen. Finish typing Y1(2) and press ENTER. Then f (2) is computed as. Enter function Return to Home Press VARS Scroll right, press 1 Press 1 for Y1 Enter Y1(2) On TI-9: After entering the function in y1, enter y1(2) to obtain f (2) =.

3 One-Sided Limits Formally, we are evaluating its. To express that x is increasing to a, we use the notation f ( x). This notation is also read as the it of f ( x) as x approaches a from the left. To express that x is decreasing to a, we use the notation f ( x). This notation + is also read as the it of f ( x) as x approaches a from the right. In either case, the actual function value at x = a is not relevant. We only care about the function values as x nears a without x ever equaling a. Example 3. Consider the following piecewise-defined function: x 2 if x < 2 f (x) = x 4 if x > 2 Evaluate f ( x) and + f ( x). What can we say about f (x )? Solution. As x increases to 2, we use the part of the function f defined for x < 2; thus, f ( x) = x2 = 4. Likewise as x decreases to 2, we use the part of the function f defined for x > 2, so + f ( x) = ( x 4) = 2. + Note that the function f is not even defined at x = 2, which does not matter in terms of evaluating the its. In order for f ( x) to exist, 1. Each of f ( x) and f ( x) must exist and be finite f ( x) must equal + f ( x). When f ( x) = f ( x) = L, then f ( x) = L also. + If f ( x) f ( x), then f ( x) does not exist. +

4 3cos(2x) if x < π /2 Example 4. Let f (x) = 6 if x = π /2 3sin(3x) if x > π /2. Evaluate x π /2 f (x) and x π /2 + f (x). What can we say about f (x)? x π /2 Solution. First, x π /2 f (x ) = 3 cos(2x ) = 3 cosπ = 3. x π /2 Next, f (x ) = x π /2 + 3sin(3x) = 3sin(3π / 2) = 3. x π/2 + Because x π /2 f (x ) = f (x ) = 3, we can say x π /2 + f ( x) = 3 also. x π /2 Note that the function value when x = π/2 is f (π / 2) = 6 The actual function value at this point does not affect the it. In this case, there is a hole in the graph, which is also called a removable discontinuity. Vertical Asymptotes As mentioned earlier, a it must be finite in order for it to be said to exist. When the value is infinite, as when approaching a vertical asymptote, then technically there is no it since the y values will continue to grow without bound. However we can still describe the behavior. Example 5. Let f (x) =. Describe the behavior of f as x approaches 4. Solution. By direct substitution of x = 4, we obtain = ±. (For a non-zero constant 0 c, then c/0 = ± which means that there is a vertical asymptote.) From the graph, we see that f (x ) = and that x 4 f (x ) = +. Because these values are infinite, these x 4 + one-sided its technically do not exist. We can still say the following: As x increases to 4, then x decreases to 4, then increases to +. decreases to ; and as

5 Some Basic Properties of Limits Assume f ( x) and g(x ) both exist. Then: 1. For any constant c, we have c f ( x) exists and c f ( x) = c f ( x). That is, the it of a constant times a function equals the constant times the it of the function. 2. ( f (x ) ± g(x )) exist and ( f (x ) ± g(x )) = f ( x) ± g( x). That is, the it of a sum is equal to the sum of the its, and the it of a difference is equal to the difference of the its. 3. ( f (x ) g(x )) exists and ( f (x ) g(x )) = f ( x) g( x). x a That is, the it of a product is equal to the product of the its. f ( x) 4. g(x ) = f ( x), provided g(x ) 0. g(x ) That is, the it of a quotient is equal to the quotient of the its.