BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 2, Problem 4RCC
To determine

To Describe: The Squeeze Theorem

Expert Solution

Explanation of Solution

Statement:

Suppose that g(x)f(x)h(x)  for all x in some open interval containing c except possibly at c itself. If limxcg(x)=L=limxch(x) then limxcf(x)=L.

Result used: Definition of limits

Let ε>0 for any δ>0 such that |f(x)L|<ε whenever 0<|xc|<δ.

By the definition of limits, since limxcg(x)=L there exist some δ1>0 such that

|g(x)L|<ε for all 0<|xc|<δ1

Proof:

Let ε>0 for any δ>0 such that |f(x)L|<ε whenever 0<|xc|<δ.

By the definition of limits, since limxcg(x)=L there exist some δ1>0 such that

|g(x)L|<ε for all 0<|xc|<δ1.

Thus, ε<g(x)L<ε for all 0<|xc|<δ1

Lε<L+g(x)L<L+ε for all 0<|xc|<δ1

Lε<g(x)<L+ε for all 0<|xc|<δ1 (1)

Similarly, since limxch(x)=L there exist some δ2>0 such that

|h(x)L|<ε for all 0<|xc|<δ2.

Thus, ε<h(x)L<ε for all 0<|xc|<δ2

Lε<L+h(x)L<L+ε for all 0<|xc|<δ2

Lε<h(x)<L+ε for all 0<|xc|<δ2 (2)

Since g(x)f(x)h(x) for all x in some open interval containing c, there exist δ3>0 such that g(x)f(x)h(x) for all 0<|xc|<δ3 (3)

Choose δ= min (δ1,δ2,δ3) then by (1),(2) and (3) equations

Lε<g(x)f(x)h(x)<L+ε for all 0<|xc|<δ

Lε<f(x)<L+ε for all 0<|xc|<δ

LεL<f(x)L<L+εL for all 0<|xc|<δ

Therefore, ε<f(x)L<ε for all 0<|xc|<δ,

So |f(x)L|<ε for all 0<|xc|<δ

By the definition of limits limxcf(x)=L_.

Example:limxsinxx=0

To prove: The limit of the function limxsin(x)x=0.

Theorem used: The Squeeze Theorem.

“If f(x)g(x)h(x) when xis near a (except possibly at a) and limxaf(x)=limxah(x)=L then limxag(x)=L.”

Proof:

First take that 1sinx1, because of the well-known properties of the sine function.

Since computing the limit as x goes to infinity, then assume that x>0.

Thus, 1xsin(x)x1x

Apply the limit that is limx(1x)limx(sin(x)x)limx(1x)

Since limx(1x)=0 and limx(1x)=0 then limx(sin(x)x)=0_.

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