Calculus! #19: Early Transcendentals 2.3

Calculating Limits Using the Limit Laws

All right! Now we’re getting to the interesting stuff. The past two chapters have really been about developing your intuition about limits. Let’s learn how limits actually work.

Here are the limit laws the book lists. In this case, c represents some constant.

 1) \lim_{x \to a} (f(x) + g(x))=\lim_{x \to a} f(x) + \lim_{x \to a} g(x)

2) \lim_{x \to a} (f(x) - g(x)) = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)
3) \lim_{x \to a} (cf(x)) = c \lim_{x \to a} f(x)
4) \lim_{x \to a} (f(x)*g(x)) = \lim_{x \to a} f(x) * \lim_{x \to a} g(x)
5) \lim_{x \to a} \dfrac{f(x)}{g(x)} = \dfrac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}, if \lim_{x \to a} g(x) \not= 0

(I tried for about an hour to get wp-latex to put a little space between the lines and it wouldn’t do it. Any ideas?)

That may seem like a lot to memorize, but it’s not bad. A lot of the formulas can be derived from others. For example, 1 and 2 are basically the same, since subtracting is just adding a negative number. And 3 and 4 are the same if you think of c as a function g(x) = c, any limit of which always approaches c.

The book goes on to derive a bunch of other laws from the initial set, which I think are all fairly obvious. There’s one principle worth noting. It’s a simple idea, but if there is any gap in your understanding of limits, it might trip you up. The book calls it the “Direct Substitution Property.” Here it is:

If f is a polynomial or a rational function, and a is in the domain of f, then \lim_{x \to a} f(x) = f(a).

That is, if you have a nice polynomial or rational function, you can just substitute in the value that x approaches to find the limit. So, for example, if your function is y=x^2, and  you’re finding the limit as x approaches 5, just substitute in 5 for x. The answer is 25. If this isn’t obvious to you, you have some gap in your understanding of limits, and need to review.

Additionally, it means if you can reduce your function to a polynomial or a rational function, you’ve basically solved the problem. So, one thing you always wanna look for is the opportunity to simplify a function. Often you’ll have a function that looks hard to analyze, until you realize the denominator can divide into the numerator once you factor the numerator.

The book gives another formula which seems so obvious I’m not sure why it’s included. I think including things that are patently clear can be bad because as a student you always assume you’re just not understanding. But, here goes:

If f(x) = g(x) when x \not= a, then \lim_{x \to a} f(x) = \lim_{x \to a} g(x) provided the limit exists.

In other words, if you have two functions that are the some when x doesn’t equal a, the limits of those functions are the same as x goes to a.

I’m going to skip through a bit of the book here, since they seem to just be restating old stuff. That said, there are some good examples of problems worked here, for the confused.

Now, we get to some good stuff.

Theorem: If f(x) is less than g(x) when x is near a (except possibly at a) and the limits of both f and g exist as x approaches a, then

\lim_{x \to a} f(x) \le \lim_{x \to a} g(x)

That is, if you have to functions that have a limit as x approaches a, and one function at that point is bigger than the other, the bigger function also has the bigger limit. This is a simple observation, but it leads to something neat:

The Squeeze Theorem:

If f(x) \le g(x) \le h(x) when x is near a (except possibly at a) and

\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L


\lim_{x \to a} g(x) = L.

Here’s what that means: Say you have two functions (f and h) that go to some value L at a particular point. And, say you have a third function, g, which you know is between f and h at that point. You therefore know that g must also go to L at that point. It’s “squeezed” between the two functions that you know surround it.

This may seem fairly simple right now, but it’ll actually pop up later as a useful fact, when you work with weird functions that are hard to analyze from a quick glance.

End of Section

This was a big section, and for good reason. Here, you get all the basic tools for solving limits, which we will use in some form throughout this entire book. It is crucial that you work the odd problems here, since this is real working knowledge stuff. You should also try to understand logically and visually why the limit laws given at the top make sense.

In the next section, we’re going to get into the mathematically rigorous definition of a limit. This is very important, but most of the time what you really are going to need is the basic tools in section 2.3. Make sure they’re second nature – pulling the constant out of a limit should be obvious to you as 2 + 2 = 4.

We’re only a few sections away  from the mighty derivative, at which point I can get back to the physics book! WOOP!

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3 Responses to Calculus! #19: Early Transcendentals 2.3

  1. Talithin says:

    “If f is a polynomial or a rational function, and a is in the domain of f, then Lim_{x\rightarrow a}f(x)=f(a)”

    I’m not sure why this is only stated to hold true for polynomials and rational functions because it’s the definition of a function being piece-wise continuous so holds for a much wider class of functions.

  2. coby says:

    “\lim_{x \to a} f(x) = \lim{x \to a} h(x) = L”
    It doesn’t really display there, but you had a bit of a latex hiccup when describing the squeeze theorem.

  3. Adam Glesser says:

    To get the extra space between lines (as well as to get the limits under the lim), try the following code:

    $latex 1) \lim\limits_{x \to a} f(x)

    2) \lim\limits_{x \to b} f(x)$

    Also, a super important point about those limit laws is that, for them to hold, you need the individual limits to exist. For instance, if f(x) = 1/x and g(x) = -1/x, they sum to 0 (except at x = 0 where they aren’t defined). Hence the limit as x approaches 0 of f(x) + g(x) is 0. But the limit as x approaches 0 is infinity for f(x) and negative infinity for g(x). The sum of infinity and negativity infinity is most decidedly not 0, however; it doesn’t exist.

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