Section 1.x: The Variety of Asymptotic Experiences

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1 calculus sin frontera Section.x: The Variety of Asymptotic Experiences We talked in class about the function y = /x when x is large. Whether you do it with a table x-value y = /x or with a graph (see Figure ), the result is the same: we write lim x x = 0, or we say the line y = 0 is a horizontal asymptote. This is all fine; but what we re going to find is that it s too limited: there are all kinds of other asymptotes, and vertical asymptote - horizontal asymptote gives only a few of all the asymptotes there are. We re going to start talking about our asymptotes differently: instead of saying "the line y = 0 is a horizontal asymptote", we ll start saying things like "y = /x is asymptotic to y = 0." Definition we say the function y = f (x) is asymptotic to the function y = g(x) if lim [ f (x) g(x)] = 0 x That is, f and g are asymptotic if they get infinitely close as x gets large. (same thing works if x ). Here s another example, one we know from the Local/Global section: y = x/(x 2 + ). In that section, we said x/(x 2 + ) /x and we checked this with charts and graphs. Now, we ll say x/(x 2 + ) is asymptotic to /x, and instead of using a chart or a graph, we ll check this algebraically: lim x x x 2 + x = lim x x 2 x(x 2 + ) x2 + x(x 2 + ) = lim x x(x 2 + ) = = 0 For my next example, I want to mix the ideas from Local/Global with those from asymptotic to. The example is y = x + /x; check out Figure 3. Here local means x near zero; try x =. for example: Figure : Global: large x When x gets large, y = /x gets small. Figure 2: Asymptotic To y = x/(x 2 + ) is asymptotic to y = /x. The graph tells you a little more: it looks like /x is always on top of x/(x 2 + ). Can you tell why, by looking at the limit computation? y = x + x =. +. =. + 0 Local/Global thinking tells me the 0 is the important part of the function, and the. unimportant. The 0 comes from the /x part of y = x + /x; this means that near zero, y = x + /x is approximately like y = /x. Figure 3: A Mixed Problem y = x + /x will have two asymptotic behaviors.

2 2 kathy davis The two are graphed together in Figure 4. You can see that near zero, the two are similar, but as x gets larger, another curve takes over. This is the global part; let s try x = 0 to count for large x : y = x + x = 0 + = Again the 0 is the important part of the function, and the. unimportant, but this time the 0 comes from the x part of y = x + /x. So for large x, y = x + /x is approximately like y = x. Let s add that to the graph; you can see the global part in Figure 5. Remember, though: while we re not gonna change our ideas about how to get local, how we do global has been changed: the politically correct way to talk is now y = x + /x is asymptotic to y = x. And we don t rely on plugging in 0 or intuitions like that: we re scientists and egnineers, and we compute things. Which, here, it isn t exactly hard: lim [ f (x) g(x)] = lim x + (x) = lim = 0 x x x x x Figure 4: Mixed Problem Near x = 0 y = x + /x is like y = /x near zero. OK, y = x + /x is asymptotic to y = x after all. Now let s put it all together: Figure 6 shows how the graph of y = x + /x sits between the two asymptotes. You ll notice the picture shows that y = x + /x is always ontop of its asymptotes. That s just algebra: when we graph, we re only looking at x > 0. Then y = x + /x > x so y = x + /x is ontop of the straight line, and, y = x + /x > /x so y = x + /x is ontop of the vertical asymptote. Figure 5: Mixed Problem Near x = y = x + /x is like y = x near infinity. Here s a couple of examples for you to think about: a) What does y = x 2 + /x look like? What it is asymptotic to? b) Same question, but with y = x 2 + /x 2. c) Same question, but with y = x + /x 2. We re not through yet; there are still more funny asymptotes. Examples a) and b) were pretty easy; you were handed the asymptotic to part by writing y = x 2 + something. For the next problem, we ll make it a little harder to see what each piece is. Figure 6: All Together y = x + /x is caught between two asymptotes.

3 calculus sin frontera 3 Our new function is y = x 3 /(x 2 + x + ). If you complete the square, you see x 2 + x + =(x + 2 ) , which means the denominator is never zero, which means f (x) doesn t have any vertical asymptotes. You see that in Figure 7, Next question: is it asymptotic to anything? How would I even figure that out? One thing to try is using that old high-school trick they tell you: you can get the limits by taking the highest power in the numerator (that d be x 3 ) and the highest power in the denominator (x 2 ). then the whole function looks like x 3 /x 2 = x. Try it; Figure 8 gives you an idea. Also, the idea it gives you is that the two aren t asymptotic; they keep their distance from each other. Oh well: good-bye high school. What we want is something like x 2 + /x. Can we write our f (x) as Figure 7: A Hard Example No vertical asymptotes; any others? g(x)+r(x)? If we could, we d get lim [ f (x) g(x)] = lim [r(x)] x x and all we d need is for r(x) 0, to get f asymptotic to g. Whatever g is. The trick here is long division. Say I let x = 0; then f (0) is 000. I want to write that as a big number plus a smaller term, and that s what a long division does: 000 = In f, g terms, if you do the long division, you get Figure 8: Try Guessing? That actually doesn t work very well. x 3 (x 2 + x + =(x )+ x 2 + x + = g(x)+r(x) just as I wanted. And, because /(x 2 + x + ) / = 0, we get x 3 /(x 2 + x + ) is asymptotic to x. Just like in Figure 9. Figure 9 also tells you a little more: the curve lies ontop of the asymptote. Again, there s an algebraic reason for that: x 3 /(x 2 + x + ) =(x )+ x 2 + x + > x = g(x) So we see graphically and algebraically that f > g, so f is ontop of g. Figure 9: Long Division! f is actually asymptotic to g(x) =x, not to g(x) =x.

4 4 kathy davis Vertical asymptotes represent places where the function doesn t exist, so a function cannot "cross" its vertical asymptote. We ve seen two kinds of asymptotes here, where again the function doesn t cross the asymptote. But that s just because we haven t done very many examples; here s a simple one. x2 + x x 2 x + 2 First you d need to long divide, and see, as in Figure 0, that g(x) =. You d need to compute f g and then compute that lim x [ f (x) g(x)] = 0. And then you d probably want to figure our why, and where, f crosses g. Of course, you don t get out of here that easily: a) What does (x 3 x 2 )/(x 2 + ) look like? What it is asymptotic to? b) Same question, but with y = x 3 /(x ). c) This is evil: y = (x 2 2x)/(x 3). Remark What is this? Why does this work? Why long division? When I write x 3 (x 2 + x + =(x )+ x 2 + x + = g(x)+r(x) Figure 0: f Crosses Its Asymptote How would you figure out where that can happen? what I m doing is writing f as two parts: the g(x) contains all the parts of f that don t go to zero at infinity, and the r(x) all the parts of f that do go to zero at infinity. Why? Because when you long divide, you get a quotient that isn t a fraction, plus a remainder that is a fraction. The quotient isn t a fraction in polynomial long division, that means there s a numerator but no denominator : look at the example, where g(x) =x. No part of that polynomial can go to zero. Now the remainder: that s a fraction. Being a fraction means that the numerator is less than the denominator. Again, in polynomial terms, that means the degree of the numerator is less than the degree of the denominator. Look at the example, where r(x) =/(x 2 + x + ); the limit will be zero. There is another way to think of g(x)+r(x): it writes f as a global part plus a local part. :)

5 calculus sin frontera 5 Level II Alert Here s a different kind of problem: show f (x) = x 2 + x is asymptotic to g(x) =x + 2. See Figure. Also, there will be cuteness. I need to compute lim x [ f (x) g(x)] = 0. Here goes: x 2 + x (x + 2 ) = x 2 + x (x + 2 ) Difference of Squares! Mine! x 2 + x +(x + 2 ) x 2 + x +(x + 2 ) = ( x 2 + x) 2 (x + 2 )2 x 2 + x +(x + 2 ) Figure : A Square Root Asymptote To show you what the evil problem might look like. = (x2 + x) (x 2 + x + 4 ) = 4 x 2 + x +(x + 2 ) x 2 + x +(x + 2 ) now the limit is easy. Side Remark And how did I guess that g(x) =x + 2?? It s a tangent line thing, again: + a + 2 a. So, x 2 + x = x 2 ( + xx 2 )= x 2 + xx 2 = x + x x( + 2 x )=x + 2 Ah, tangent lines: can t live without em, can t live with em. Figure 2: Chibi Nadie I threatened cuteness. Points for making good on threat?