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Chapter 1: Q. 97 (page 137)

In the reading, we used the Squeeze Theorem to prove that limh鈫�0sinh=0and limh鈫�0cosh=1. Use these facts, the sum identity for cosine, and limit rules to prove thatrole="math" localid="1648152969592" f(x)=cosx is continuous everywhere.

Short Answer

Expert verified

Ans: limx鈫�ccosx=f(c)

Step by step solution

01

Step 1. Given Information:

The strategy is to prove using Squeeze theorem, that f(x)=cosxis continuous everywhere.

02

Step 2. Prove:

Givenc鈭�鈩�,limx鈫�cf(x)=limx鈫�ccosx=limh鈫�0cos(c+h)=limh鈫�0(cos(c)cos(h)-sin(c)sin(h))=cos(c)(limh鈫�0cosh)-sin(c)(limh鈫�0sinh)=cosc(1)-sinc(0)=cosc-0=cosc=f(c)

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Most popular questions from this chapter

Write delta-epsilon proofs for each of the limit statements limx鈫�cfx=Lin Exercises 47-60.

limx鈫�03x2+1=1.

State what it means for a functionf to be continuous at a point x = c, in terms of the delta鈥揺psilon definition of limit.

Use algebra to find the largest possible value of 未 or smallest possible value of N that makes each implication true. Then verify and support your answers with labeled graphs.

Ifx>N, then1x2鈭�0<0.001

Sketch a labeled graph of a function that fails to satisfy the hypothesis of the Intermediate Value Theorem, and illustrate on your graph that the conclusion of the Intermediate Value Theorem does not necessarily hold.

Write delta-epsilon proofs for each of the limit statements limx鈫�cfx=Lin Exercises 47鈥�60.

limx鈫�1x2-1x-1=2
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