EXAMPLE 4 Investigate the following limit. lim sin 3T x - 0 SOLUTION Again the function f(x) = sin(3m/x) is undefined at 0. Evaluating the function for some small values of x, we get f(1) = sin(3It) = ( 2 ) = sin(6TT) = 0 (= sin(91) = ( 4 ) = sin( IT = 0 f(0.1) = sin(30m) = 0 f(0.01) = sin(300m) = Similarly, f(0.001) = f(0.0001) = 0. On the basis of this information we might be tempted to guess that lim sin 31 x - C but this time our guess is wrong. Note that although f(1) = sin(3nn) = for any integer n, it is also true that f(x) = 1 for infinitely many values of x that approach 0. You can see this from the graph of f given in the figure. The compressed lines near the y-axis indicate that the values of f(x) oscillate between 1 and -1 infinitely often as x approaches 0. (See this exercise. ) Since the values of f(x) do not approach a fixed number as x approaches 0, lim sin( 3TC does not exist. X -sin(x) EXAMPLE 3 Guess the value of the limit below. X X lim sin (x) +1.0 0.84147098 x - 0 X +0.5 0.95885108 SOLUTION The function f(x) = (sin(x))/x is not defined when x = . Using a calculator ( and remembering that, if x E R, sin(x) means the sine of the angle whose radian measure is x), we +0.4 0.97354586 construct a table of values correct to eight decimal places. From the table and the graph in the figure 10.3 0.98506736 we guess that +0.2 0.99334665 lim sin(x) x - 0 X +0. 1 0.99833417 This guess is in fact correct, as will be proved in Chapter 3 using a geometric argument. +0.05 0.99958339 10.01 0.99998333 sin x y= 10.005 0.99999583 X +0.001 0.99999983 Video Example () -1 0 XEXAMPLE 7 The graph of a function g is shown in the figure. Use it to state the values (if they exist) of the following: 3 (a) lim g(x) (b) lim g(x) (c) lim g(x) x - 2- x - 2+ x - 2 (d) lim g(x) (e) lim g(x) (f) lim_g(x). x - 5- x - 5+ X - SOLUTION From the graph we see that the values of g(x) approach as x approaches 2 from the left, but they approach as x approaches 2 from the right. Therefore X -2 2 6 (a) lim g(x) = and (b) lim g(x) = Video Example () x - 2 x - 2+ (c) Since the left and right limits are different, we conclude that the limit as x approaches 2 of g(x) does not exist. The graph also shows that (d) lim g(x) = and (e) lim g(x) = X - 5 x - 5+ (f) This time, the left and right limits are the same and so, by this theorem, we have lim_g(x) = x - 5 Despite this fact, notice that g(5) # 3