Topic 1.5 key terms
- limit laws
- If lim_(x→a) f(x) = L and lim_(x→a) g(x) = M (finite), then f ± g → L ± M, kf → kL, fg → LM, and f/g → L/M provided M ≠ 0, all as x → a.
- direct substitution
- For a polynomial, or a rational function whose denominator doesn't approach 0 at a, the limit as x → a equals the value at a, because the limit laws split it into pieces that each approach their value.
- limit of a composite
- If g(x) → M and the outside function F is continuous at M, then F(g(x)) → F(M).
- indeterminate form 0/0
- When substitution makes the top and the bottom both 0, the limit isn't determined yet: it could be any number or not exist. Rewrite first; never write a limit equal to 0/0.
- factor and cancel
- Factor the top and the bottom and cancel a common factor that is nonzero for x ≠ a; the new expression equals the old one near a, so it has the same limit.
- same nearby outputs, same limit
- If f(x) = g(x) for every x near a (except possibly at a), then lim f(x) = lim g(x) as x → a. This is why rewriting is allowed.
- conjugate
- For √A − B, the conjugate is √A + B. Multiplying the top and bottom by it (a fraction equal to 1) turns the difference of squares into A − B².
- Pythagorean rewrite
- sin²θ = 1 − cos²θ = (1 − cos θ)(1 + cos θ), which exposes a factor that cancels.
- one-sided limits (analytic)
- Find the limit from the left with the formula that holds just left of a, and from the right with the one just right of a. The two-sided limit exists only if they agree.
Transcript
1,057 words · 3 parts
Three seconds
This is the Leaning Tower of Pisa, about forty minutes before sunset. The story goes that a young professor named Galileo once climbed it and dropped balls off the top, just to see how they’d fall.
Historians doubt it ever happened, so tonight we’re doing it for him, with an iron ball on a magnet forty-four point one meters above a pit of sand. Let’s switch off the magnet and count the seconds.
That took exactly three seconds, and what I’d really like to know is how fast the ball was moving at the very instant it hit the sand. Hold onto a guess, in kilometers an hour if that’s easier.
You already know that a speed at one instant is a limit, and so far we’ve only been able to creep up on one. Let’s do that with a live readout, how far the ball still has to fall divided by how long it has left.
As the ball gets closer, the average climbs past twenty-nine and keeps creeping upward. Each smaller time window gives us a better estimate, but we still haven’t reached the instant itself. Then, right at the instant we care about, it breaks. There’s no distance left and no time left, and if you type zero divided by zero into your own calculator, you get an error too.
By the end of this lesson, you’ll be able to get exact answers right where your calculator gives up. When a fraction falls apart into zero over zero, you’ll know how to rewrite it and pull out the real answer with algebra. The first exact answer we’ll get is the one this readout couldn’t give us, the speed of this ball at the instant it hits.
Nothing left to divide
I’ve parked the ball one second before it lands, twenty-four and a half meters up. It has exactly one second to cover those meters, so its average for that last second is twenty-four point five meters per second.
Now it’s your turn to drive. Drag the time left down until the ball has a hundredth of a second or less to go, without letting it hit, and keep an eye on the readout.
Let’s turn that readout into a formula, because a formula is something algebra can work on. In t seconds a dropped ball falls four point nine t squared meters, so ours still has forty-four point one minus four point nine t squared to go. It lands at three seconds, so it has three minus t seconds left, and the readout is the height left divided by the time left.
Every number in your table came out of this one fraction, and they’re clearly heading somewhere near twenty-nine point four. A table only ever gets close, though. Plugging in three itself makes the top forty-four point one minus forty-four point one and the bottom three minus three, so substitution gives zero over zero.
That leaves us two questions. When does plugging in really give you the limit, and what do you do when it hands you zero over zero? Let’s start with the easy one.
When plugging in just works
Let’s warm up with the ball’s height itself, and ask where it’s heading as the time, t, gets close to two seconds. You’d probably just plug in two, which gives forty-four point one minus four point nine times four, twenty-four point five meters, and that’s right.
A limit tells us where the height is heading, so why does plugging in give us the right answer here? The reason is a set of rules called the limit laws. The limit of a difference is the difference of the limits, and a constant like four point nine just comes along for the ride. The limit of a product is the product of the limits.
As t heads to two, t times t heads to four, so every piece heads right where plugging in says it does. That’s why plugging in works for any polynomial. In general, if f heads to L and g heads to M, then f plus g heads to L plus M and f times g heads to L times M. Their quotient heads to L over M, as long as M isn’t zero.
The exam loves to skip the formulas and just give you limits, like f heading to three and g heading to negative two as x heads to four. Two f minus g heads to six minus negative two, which is eight, and f times g heads to negative six. We’re allowed to divide too, because g’s limit isn’t zero, so f over g heads to negative three halves.
With a function wrapped around the outside, work from the inside out. Inside, f squared plus seven heads to sixteen, and the square root has no break at sixteen, so the whole thing heads to four. That’s the composite law’s fine print. You can pass the limit through the outside function as long as it’s continuous, with no break, right at the value the inside is heading to.
Let’s see you put the laws together. With those same two limits, what does f squared plus g, all over f plus g, head to as x heads to four?
Now let’s go back to the ball’s readout. At two seconds its top heads to twenty-four point five and its bottom heads to one, so the quotient law gives twenty-four point five meters per second.
At three seconds the bottom heads to zero, and the quotient law tells you up front that it can’t help when the bottom’s limit is zero. Zero over zero is a signal to stop and rewrite. Look at x over x, five x over x, and x squared over x as x heads to zero, because all three give zero over zero when you plug in.
Look at all three. Do they head to the same number? Once you simplify them, they head to one, five and zero, three different answers, because it all depends on how fast the top shrinks compared with the bottom. That’s why mathematicians call zero over zero an indeterminate form, since on its own it doesn’t settle the answer.
Before we crack one open, let’s make sure you can spot the signal. Which of these limits, all as x heads to three, can you not find just by plugging in?