f(x) = lim x 0 + x = lim f(x) =

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1 Infinite Limits Having discussed in detail its as x ±, we would like to discuss in more detail its where f(x) ±. Once again we would like to emphasize that ± are not numbers, so if we write f(x) = we are not saying that the it exists. What we are actually saying is that the it does not exist, and it does not exist because the values of the function f(x) grow without bound as x x 0. Similarly, f(x) = means that the it as x x 0 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 its 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. Example Find and What can you conclude about 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 x 0 x = x 0 When we look at x 0 + the values of x are becoming very small, but remain positive. Thus, x 0 + x = Since the one-sided its are different, we write that x 0 + does not exist. Note that x even if both f(x) = f(x) = x 0 x 0 + the it would still not exist, but we would write f(x) = x 0 to signify the it does not exist because as x 0 from both sides the values of f(x) grow without bound. Example 2 Find x 4 (x + 4) 4 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

2 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 it does not exist). We write x 4 (x + 4) 4 = to signify the function values grow without bound as x 4 from both sides. Example 3 Find x 3 (x 3) 4 Solution The first thing for us to do is simplify this fraction as much as possible. If we factor from the numerator, we find that x 3 (x 3) = (x 3) 4 x 3 (x 3) = 4 x 3 (x 3) 3 Note that above we were able to cancel the factor of (x 3) because in looking at the it as x 3, we know that x 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 its rather than trying to calculuate a two-sided it directly (because they will have different signs, so unless they are both 0, the two-sided it will not exist). As x 3 the denominator is negative, so the entire fraction is positive (because there is a 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 x 3 (x 3) 4 = x 3 = and (x 3) 3 x 3 + (x 3) 4 = x 3 + (x 3) 3 = Since our function has no consistent behavior as x 3, all we can say is that the it does not exist. Having considered infinite its intuitively, let us introduce the formal definition.

3 Definition: Infinite Limits We write f(x) = if for every number B > 0 there exists a corresponding δ > 0 such that for all x with 0 < x x 0 < δ we have f(x) > B Similarly, we write f(x) = if for every number B < 0 there exists a corresponding δ > 0 such that for all x with 0 < x x 0 < δ we have f(x) < B Example 4 Prove that x = 2 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 /x 2 > B. Manipulating this relationship, we find that we require B > x2 which means that we need x < B x 0 Thus, we need to choose 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 it 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 x a x a f(x) = ± or f(x) = ± + 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). Example 5 Find the horizontal and vertical asymptotes of f(x) = x + 2 Solution Horizontal asymptotes correspond to constant values in the its as x and x, and x x + 2 = and x x + 2 = so the only horizontal asymptote is the line y =. To look for vertical asymptotes, we want to look at places where the denominator of the function approaches 0; in this case we need

4 to consider x 2. It is enough to begin with just a one-sided it, because we do not require a two-sided it to approach to classify a vertical line as an asymptote. Since x 2 + 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 =, 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 /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 tan(x) = sin(x) cos(x) in order to progress in this problem. Since sin(x) is ± 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 2 + nπ Example 7 Find all asymptotes of f(x) = 3x 2 5 Solution In this case the denominator has zeroes at both x = /3 and x = 2. However, we cannot immediately conclude that x = /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 0 0 (the only way we can be certain that we have found a vertical asymptote is by evaluating the appropriate it). If we look as x 2 we find that 3x 2 5 = (3x + )() = 3x + = 7 which is very far from a it 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 it which does not exist, because x /3 x /3 + 3x + 3x 2 5 = x /3 3x + = and x /3 3x + =

5 Nevertheless, we don t require the two-sided it approach ± for a vertical asymptote to existr; in conclusion, we have a vertical asymptote at x = /3 but not at x = 2. We also find that x ± f(x) = 0 so we have a horizontal asymptote of y = 0.