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Chapter 1: Q. 82 (page 122)

Write a delta鈥揺psilon proof that shows that the function f(x)=2x+1 is continuous at x=5. (This exercise depends on Section 1.3.)

Short Answer

Expert verified

Hence we proved limx鈫�5(2x+1)=11.

Step by step solution

01

Step 1. Given Information 

We are given the function f(x)=2x+1 is continuous atx=5 and we need to write a delta鈥揺psilon proof.

02

Step 2. Delta–epsilon proof  

f(x)=2x+1c=5L=2(5)+1=10+1=11

For 蔚>0,未=蔚2,

If 0<x-5<未we have,

2x+1-11=2x-10=2x-5<2未=2蔚2=蔚鈭�limx鈫�52x+1=11

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