Essential Knowledge 1.1C2

Essential Knowledge 1.1C2 Students will know that the limit of a function may be found by using algebraic manipulation, alternate forms of trigonometric functions, or the squeeze theorem.

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The AP test will most likely give you limit questions that aren't so easy. Most likely, the limit will be indeterminate, and you'll have to find a way to mess with it to get a better answer. You will learn many strategies for taking an indeterminate answer and making it a real number. The college board identifies four main strategies, one which you can't learn until later.

I. Algebraic Manipulation

II. Alternate Forms of Trigonometric Functions

III. Squeeze Theorem (Sandwich Theorem)

IV. L'Hôpital's Rule

Let's look at some examples for each of these

I. Algebraic Manipulation

Example 1 (Multiplying by a clever form of 1)

Using direct substitution, we would get

which is indeterminate

Let's multiply by a clever form of 1. Usually you look and see what the largest power of the variable is. The largest power of x in this problem is x2. Then you make a fraction with the same numerator and denominator like so:

We can continue simplifying this and we will get a limit we can use

II. Alternate Forms of Trigonometric Functions

Example 1

Using direct substitution, we would get

which is indeterminate

Let's use trigonometric identities. Well, can't do much with sinx. But tanx is the same as sinx/cosx. So lets try plugging that in

We can continue simplifying this and we will get a limit we can use

Example 2 (Factoring and Simplifying)

III. Squeeze Theorem (Sandwich Theorem)

Squeeze Theorem actually requires some creativity. Usually it's a last resort. Basically, if you can squeeze your function between two other functions that meet at a point, then your function must also meet at that point. An example makes the squeeze theorem much more clear, but here is the theorem

Given g(x) ≤ f(x) ≤ h(x), if , then

Example 1

Using direct substitution, we would get

which is a problem because sin∞ isn't defined.

But don't give up. Let's look at a graph.

You can see that x2sin(1/x) never escapes between -x2 and x2. This is perfect, because -x2 and x2 both equal 0 when x = 0. Therefore, since x2sin(1/x) can't be less than 0 or bigger than 0 when x = 0, x2sin(1/x) must equal 0 when x = 0. Using the squeeze theorem, would look like this

-x2 ≤ x2sin(1/x) ≤ x2, and

We know sine cannot produce values less than -1 or greater than 1. That means the biggest the function can be is x2(1). The smallest the function can be is x2(-1). Lets put these on the graph.

therefore

= 0

IV. L'Hôpital's Rule

You'll learn this one later.