Infinite Limits 6 Examples

Infinite Limits Having discussed in detail limits as x → ±∞, we would like to discuss in more detail limits where f (x) → ±∞. Once again we would like to emphasize that ±∞ are not numbers, so if we write lim f (x) = ∞ x→x0 we are not saying that the limit exists. What we are actually saying is that the limit does not exist, and it does not exist because the values of the function f (x) grow without bound as x → x0 . Similarly, lim f (x) = −∞ x→x0 means that the limit as x → x0 of f (x) does not exist because the function values decrease without bound (become arbitrarily large in magnitude, and negative in sign). Throughout this discussion one-sided limits will be particularly useful to us, because it is quite common that a function may approach ∞ from one side of a point, and −∞ from the other side of the point. 1 1 1 and lim+ What can you conclude about lim ? x→0 x x→0 x x→0 x Solution When x → 0− we are considering negative x values, which are becoming smaller and smaller, approaching 0. Since the magnitude of the denominator becomes very small, the magnitude of the fraction becomes very large, growing without bound. Since the x values are negative, it follows that 1 lim− = −∞ x→0 x When we look at x → 0+ the values of x are becoming very small, but remain positive. Thus, 1 lim+ = ∞ x→0 x Example 1 Find lim− Since the one-sided limits are different, we write that limx→0+ even if both lim− f (x) = lim+ f (x) = ∞ x→0 1 x does not exist. Note that x→0 the limit would still not exist, but we would write lim f (x) = ∞ x→0 to signify the limit does not exist because as x → 0 from both sides the values of f (x) grow without bound. Example 2 Find 1 x→−4 (x + 4)4 lim Solution In analyzing this function, the first thing to notice is that we have a fourth power in the denominator, (x + 4)4 . Since (x + 4) is raised to an even power, the denominator is always non-negative (it is 0 when x = −4 and positive otherwise). Since the numerator is always positive as well, it means that this function is always non-negative. As x → −4− and x → 4+ the denominator approaches 0, which means that the values of the fraction grow without bound (so the limit does not exist). We write 1 =∞ x→−4 (x + 4)4 lim to signify the function values grow without bound as x → −4 from both sides. Example 3 Find 3−x x→3 (x − 3)4 lim Solution The first thing for us to do is simplify this fraction as much as possible. If we factor −1 from the numerator, we find that −(x − 3) −1 3−x = lim = lim 4 4 x→3 (x − 3) x→3 (x − 3)3 x→3 (x − 3) lim Note that above we were able to cancel the factor of (x − 3) because in looking at the limit as x → 3, we know that x 6= 3, so the factor will never be 0 (we, of course, cannot divide by 0). In this case, unlike the previous one, we have an odd power in the denominator; that is, (x − 3) is raised to the third power. Because of this, we will have (x − 3)3 < 0 when x < 3 and (x − 3)3 > 0 when x > 3 Since the sign of the denominator is different depending on what side x → 3 from, we should look at the one-sided limits rather than trying to calculuate a two-sided limit directly (because they will have different signs, so unless they are both 0, the two-sided limit will not exist). As x → 3− the denominator is negative, so the entire fraction is positive (because there is a −1 in the numerator). As x → 3+ the denonominator is always positive, so the entire function is negative. In both cases the denominator approaches 0, so we find that lim− x→3 −1 3−x = lim− = ∞ and 4 x→3 (x − 3)3 (x − 3) lim+ x→3 3−x −1 = lim+ = −∞ 4 x→3 (x − 3)3 (x − 3) Since our function has no consistent behavior as x → 3, all we can say is that the limit does not exist. Having considered infinite limits intuitively, let us introduce the formal definition. Definition: Infinite Limits We write lim f (x) = ∞ x→x0 if for every number B > 0 there exists a corresponding δ > 0 such that for all x with 0 < |x − x0 | < δ we have f (x) > B Similarly, we write lim f (x) = −∞ x→x0 if for every number −B < 0 there exists a corresponding δ > 0 such that for all x with 0 < |x − x0 | < δ we have f (x) < −B 1 Example 4 Prove that lim 2 = ∞ x→0 x Solution We will start as previously, but now consider B > 0, arbitrary. We need to find δ > 0 so that for all x with 0 < |x−0| < δ we have 1/x2 > B. Manipulating this relationship, we find that we require 1 > x2 B which means that we need 1 |x| < √ B Thus, we need to choose 1 δ≤√ B For arbitrary x with |x| < δ we have f (x) > B; The conclusion follows. Just as we defined horizontal asymptotes by looking at functions with a finite limit as x → ±∞, we can define vertical asymptotes, which occur when f (x) → ±∞. More precisely, the line x = a is a vertical asymptote of the graph of a function if either lim f (x) = ±∞ or x→a+ lim f (x) = ±∞ x→a− This means that if for any point a the magnitude of f (x) grows without bound (so f (x) → ±∞) as x → a from either side, then x = a is a vertical asymptote of the graph of f (x). x+3 x+2 Solution Horizontal asymptotes correspond to constant values in the limits as x → ∞ and x → −∞, and x+3 x+3 lim = 1 and lim =1 x→∞ x + 2 x→∞ x + 2 so the only horizontal asymptote is the line y = 1. To look for vertical asymptotes, we want to look at places where the denominator of the function approaches 0; in this case we need Example 5 Find the horizontal and vertical asymptotes of f (x) = to consider x → −2. It is enough to begin with just a one-sided limit, because we do not require a two-sided limit to approach ∞ to classify a vertical line as an asymptote. Since lim + x→−2 x+3 =∞ x+2 We conclude that x = −2 is a vertical asymptote. Since the denominator does not approach 0 in any other places, there are no other vertical asymptotes. For this function the horizontal and vertical asymptotes have a nice graphical interpretation. If we draw the lines corresponding to y = 2 and x = 1, we know the behavior as x → ±∞ and x → 2 will be in some way represented by this asymptotes. In fact, our function looks exactly like 1/x, if we simply were to shift the x and y axes to be these two asymptotes. Example 6 Find the vertical asymptotes of f (x) = tan(x). Solution Since vertical asymptotes correspond to places where denominator approaches 0, we need to rewrite sin(x) tan(x) = cos(x) in order to progress in this problem. Since sin(x) is ±1 when cos(x) = 0, it follows that every time cos(x) has a zero, tan(x) will have a vertical asymptote. These occur every π radians, and more precisely, tan(x) has a vertical asymptote for x= π + nπ 2 x−2 3x2 − 5x − 2 Solution In this case the denominator has zeroes at both x = −1/3 and x = 2. However, we cannot immediately conclude that x = −1/3 and x = 2 are vertical asymptotes, because if the numerator is also 0 in one of these places, we will have an indeterminate form of 00 (the only way we can be certain that we have found a vertical asymptote is by evaluating the appropriate limit). If we look as x → 2 we find that Example 7 Find all asymptotes of f (x) = lim x→2 3x2 x−2 x−2 1 1 = lim = lim = x→2 x→2 − 5x − 2 (3x + 1)(x − 2) 3x + 1 7 which is very far from a limit which approaches ∞. Thus, even though we seemed to have division by 0 in the denominator at x = 2, the numerator was also 0, so we had an undefined expression, of which we could not be certain if it would grow without bound (in this case it did not, but in others it might!). Looking at the other limit x−2 1 = lim x→−1/3 3x2 − 5x − 2 x→−1/3 3x + 1 lim which does not exist, because lim x→−1/3+ 1 = ∞ and 3x + 1 lim x→−1/3− 1 = −∞ 3x + 1 Nevertheless, we don’t require the two-sided limit approach ±∞ for a vertical asymptote to existr; in conclusion, we have a vertical asymptote at x = −1/3 but not at x = 2. We also find that lim f (x) = 0 x→±∞ so we have a horizontal asymptote of y = 0.