MATH 12002 - CALCULUS I 1.4: Calculating Limits - More Examples - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §1.4: Calculating Limits - More Examples

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 7

Limits & Radicals

Example 1

Evaluate lim

x→3

√x + 1 − 2 x − 3 . Observe first that when x = 3, both the numerator and denominator are 0. This indicates that x − 3 is a factor of the numerator. To use “cancellation” to evaluate the limit, we need the other factor. When evaluating limits involving radicals, it is often helpful to “rationalize” the numerator or denominator. In this case, we multiply the numerator and denominator by the conjugate, √x + 1 + 2, of the numerator and use the fact that (a − b)(a + b) = a2 − b2 to obtain (√x + 1 − 2)(√x + 1 + 2) = (√x + 1)2 − 22 = (x + 1) − 4 = x − 3. [Continued →]

D.L. White (Kent State University) 2 / 7

Limits & Radicals

We therefore have lim

x→3

√x + 1 − 2 x − 3 = lim

x→3

(√x + 1 − 2)(√x + 1 + 2) (x − 3)(√x + 1 + 2) = lim

x→3

(x + 1) − 4 (x − 3)(√x + 1 + 2) = lim

x→3

x − 3 (x − 3)(√x + 1 + 2) = lim

x→3

1 √x + 1 + 2, since x = 3 = 1 √3 + 1 + 2, by Limit Laws = 1 √ 4 + 2 = 1 2 + 2 = 1 4.

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Limits & Sums

We know that if lim

x→a f (x) and lim x→a g(x) both exist, then

lim

x→a[f (x) + g(x)] = lim x→a f (x) + lim x→a g(x).

But it is also possible for lim

x→a[f (x) + g(x)] to exist

even if lim

x→a f (x) and lim x→a g(x) do not.

Example 2

Evaluate lim

x→1

• 3

x − 1 + x2 − 2x + 7 (x − 1)(x − 3)

• .

Note that neither lim

x→1

3 x − 1 nor lim

x→1

x2 − 2x + 7 (x − 1)(x − 3) exists, because both denominators are 0 when x = 1, while neither numerator is 0. To evaluate the limit, we will need to combine the fractional expressions. [Continued →]

D.L. White (Kent State University) 4 / 7

Limits & Sums

We have lim

x→1

• 3

x − 1 + x2 − 2x + 7 (x − 1)(x − 3)

• = lim

x→1

• 3(x − 3)

(x − 1)(x − 3) + x2 − 2x + 7 (x − 1)(x − 3)

• = lim

x→1

3(x − 3) + x2 − 2x + 7 (x − 1)(x − 3) = lim

x→1

3x − 9 + x2 − 2x + 7 (x − 1)(x − 3) = lim

x→1

x2 + x − 2 (x − 1)(x − 3) = lim

x→1

(x − 1)(x + 2) (x − 1)(x − 3) = lim

x→1

x + 2 x − 3, since x = 1 = 1 + 2 1 − 3 = 3 −2 = −3 2.

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Squeeze Theorem

Some limits may be difficult to compute directly, but may be closely related to other limits that are easy to compute. In these cases, the Squeeze Theorem may be useful.

Squeeze Theorem

If f (x) g(x) h(x) for all x in an open interval around x = a (except possibly at x = a), and lim

x→a f (x) = lim x→a h(x) = L,

then lim

x→a g(x) = L.

Example 3

Evaluate lim

x→0(sin2 x) cos 1 x .

Observe that as x → 0, sin2 x → sin2 0 = 0, but lim

x→0 cos 1 x does not exist.

Therefore, we cannot use limit laws to evaluate the limit of the product. We can, however, use the Squeeze Theorem. [Continued →]

D.L. White (Kent State University) 6 / 7

Squeeze Theorem

We know that −1 cos 1 x 1. Since sin2 x 0 for all x, we can multiply each part of this inequality by sin2 x to obtain − sin2 x (sin2 x) cos 1 x sin2 x. As we observed above, lim

x→0 sin2 x = 0, and so lim x→0(− sin2 x) = 0.

Therefore, (sin2 x) cos 1

x is squeezed between − sin2 x and sin2 x,

both of which approach 0 as x approaches 0. Hence the Squeeze Theorem implies lim

x→0(sin2 x) cos 1 x = 0.

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