MATH 12002 - CALCULUS I 1.4: Calculating Limits Professor Donald L. - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §1.4: Calculating Limits

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 6

Cancellation

One of our Limit Laws states that if f (x) and g(x) are functions such that lim

x→af (x) and lim x→ag(x) both exist, and lim x→ag(x) = 0, then

lim

x→a

f (x) g(x) = lim

x→af (x)

lim

x→ag(x).

But what if lim

x→ag(x) = 0?

We will consider the situation where lim

x→ag(x) = 0 and lim x→af (x) = 0

in §1.6. For now, we will consider the situation where both lim

x→ag(x) = 0 and lim x→af (x) = 0.

D.L. White (Kent State University) 2 / 6

Cancellation

If f (x) and g(x) happen to be polynomials and lim

x→ag(x) = 0 and

lim

x→af (x) = 0, then f (x) and g(x) will have the common factor x − a,

which we will be able to “cancel” in order to obtain a limit we may be able to compute. Recall our example from §1.3:

Example

Find lim

x→2 x2−4 x−2 . We have

lim

x→2

x2 − 4 x − 2 = lim

x→2

(x − 2)(x + 2) x − 2 = lim

x→2(x + 2)

= 2 + 2 = 4. But why are we allowed to “cancel” the x − 2 factors in this equation?

D.L. White (Kent State University) 3 / 6

Cancellation

In general, if r = 0, then r

r = 1, and so if s and t = 0 are numbers, then

rs rt = r r · s t = 1 · s t = s t . If r = 0, this equation is not valid, however. In particular, if f (x) = x2 − 4 x − 2 = (x − 2)(x + 2) x − 2 and g(x) = x + 2, then f (2) is undefined, while g(2) = 4. Therefore, as functions of x, f (x) = g(x). They are not the same function! So why are lim

x→2f (x) and lim x→2g(x) equal?

D.L. White (Kent State University) 4 / 6

Cancellation

Their graphs may give us a clue:

• ❜
• r

f (x) = x2 − 4 x − 2 g(x) = x + 2 Notice that the only point where the functions differ is at x = 2.

D.L. White (Kent State University) 5 / 6

Cancellation

Let’s look at this situation algebraically. For all x = 2 we have x − 2 = 0, and so x−2

x−2 = 1.

Hence, as before, for x = 2, we have f (x) = (x − 2)(x + 2) x − 2 = x − 2 x − 2 · (x + 2) = 1 · (x + 2) = (x + 2) = g(x). So for all x except x = 2, we have f (x) = g(x). Now recall our definition of limit:

Definition

Let y = f (x) be a function and let a and L be numbers. We say that the limit of f as x approaches a is L if y can be made arbitrarily close to L by taking x close enough to a, but not equal to a. What happens at x = a is not relevant! Therefore, since f (x) = g(x) for all x = 2 and x = 2 is not considered when computing the limit as x → 2, we have lim

x→2f (x) = lim x→2g(x).

D.L. White (Kent State University) 6 / 6