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Chapter 10: Q. 90 (page 777)

Prove the first part of Theorem 1.31(a): If k>0, then limx鈫�鈭�xk=鈭�. (Hint: Given M>0, choose N=M1/k. Then show that for x>N it must follow that xk>M.)

Short Answer

Expert verified

It is proved that If k>0, then limx鈫�鈭�xk=鈭�.

Step by step solution

01

Step 1. Given Information 

We are given that k>0,then limx鈫�鈭�xk=鈭�.

02

Step 2. Proving the statement 

Consider a positive number M and N=M1k, where k>0.

There is a real value of x which is greater than N that is x>N.

Substitute x>Nin the equation N=M1kand simplify,

x>N=M1k

Take k th power of both sides of an inequality x>M1k,

xk>M1kkxk>M1k路kxk>M

The value of M is greater than 0 and for every such M, there exists an x such that xk>M.

Hence Proved.

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