Chapter 10: Q. 90 (page 777)

Prove the first part of Theorem $1.31$ (a): If $k > 0$ , then $\lim_{x \to \infty} x^{k} = \infty$ . (Hint: Given $M > 0$ , choose $N = M^{1 / k}$ . Then show that for $x > N$ it must follow that $x^{k} > M$ .)

Short Answer

Expert verified

It is proved that If $k > 0$ , then $\lim_{x \to \infty} x^{k} = \infty$ .

Step by step solution

Step 1. Given Information

We are given that $k > 0,$ then $\lim_{x \to \infty} x^{k} = \infty$ .

Step 2. Proving the statement

Consider a positive number M and $N = M^{(\frac{1}{k})}$ , where $k > 0$ .

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

Substitute $x > N$ in the equation $N = M^{(\frac{1}{k})}$ and simplify,

$x > N = M^{(\frac{1}{k})}$

Take k ^th power of both sides of an inequality $x > M^{(\frac{1}{k})}$ ,

$x^{k} > {(M^{(\frac{1}{k})})}^{k} x^{k} > M^{\frac{1}{k} \cdot k} x^{k} > M$

The value of M is greater than $0$ and for every such M, there exists an x such that $x^{k} > M$ .

Hence Proved.

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Prove the first part of Theorem $1.31$ (a): If $k > 0$ , then $\lim_{x \to \infty} x^{k} = \infty$ . (Hint: Given $M > 0$ , choose $N = M^{1 / k}$ . Then show that for $x > N$ it must follow that $x^{k} > M$ .)

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Proving the statement

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

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Proving the statement

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Calculus

Applied Mathematics

Theoretical and Mathematical Physics

Probability and Statistics

Pure Maths

Study anywhere. Anytime. Across all devices.

Prove the first part of Theorem $1.31$ (a): If $k > 0$ , then $\lim_{x \to \infty} x^{k} = \infty$ . (Hint: Given $M > 0$ , choose $N = M^{1 / k}$ . Then show that for $x > N$ it must follow that $x^{k} > M$ .)