SECTION 14.3 One-sided Limits; Continuity 957 Figure 11 R x x x x 2 6 8 2 ( ) = − − + 1 2 21 22 2 x 5 4 4 6 x y 2 1 2 If x 4 > and x gets closer to 4, the value of x 1 4 − is positive and becomes unbounded; that is, R x lim . x 4 ( ) = ∞ → + Since R x( ) →∞ for x close to 4, the graph of R has a vertical asymptote at x 4. = (c) It is easiest to graph R by observing that x if 2,then ≠ R x x x x x 2 2 4 1 4 ( ) ( )( ) = − − − = − Therefore, the graph of R is the graph of y x 1 = shifted to the right 4 units with a hole at 2, 1 2 . ( ) − See Figure 11. Now Work PROBLEM 73 The exponential, logarithmic, sine, and cosine functions are continuous at every number in their domain. The tangent, cotangent, secant, and cosecant functions are continuous except at numbers for which they are not defined, where vertical asymptotes occur.The square root function and absolute value function are continuous at every number in their domain. The function f x x int ( ) ( ) = is continuous except for x an = integer, where a jump occurs in the graph. Piecewise-defined functions require special attention. Figure 12 f x x x x x x x if 0 1 if 0 2 5 if 2 5 2 ( ) = ≤ + < < − ≤ ≤ ⎧ ⎨ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ x y 2 3 4 22 22 2 (0, 0) (2, 3) 4 6 Solution The “pieces” of f —that is, y x y x , 1, 2 = = + and y x 5 , = − are each continuous for every number since they are polynomials. In other words, when we graph the pieces, we will not lift our pencil. When we graph the function f, however, we have to be careful, because the pieces change at x 0 = and at x 2. = So the numbers we need to investigate for f are x 0 = and x 2. = • x f At 0: 0 0 0 2 ( ) = = = Determining Where a Piecewise-defined Function Is Continuous Determine the numbers at which the following function is continuous. f x x x x x x x if 0 1 if 0 2 5 if 2 5 2 ( ) = ≤ + < < − ≤ ≤ ⎧ ⎨ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ EXAMPLE 3 f x x f x x lim lim 0 lim lim 1 1 x x x x 0 0 2 0 0 ( ) ( ) ( ) = = = + = → → → → − − + + f x x x if 0 2 ( ) = ≤ f x x x 1 if 0 2 ( ) = + < < Since f x f lim 0 , x 0 ( ) ( ) ≠ → + f is not continuous at x 0. = • x f At 2: 2 5 2 3 ( ) = = − = f x x f x x lim lim 1 3 lim lim 5 3 x x x x 2 2 2 2 ( ) ( ) ( ) ( ) = + = = − = → → → → − − + + So f is continuous at x 2. = The graph of f, given in Figure 12, demonstrates the conclusions drawn above. Now Work PROBLEMS 53 AND 61
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