Q. 90 - Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1. Given Information

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

Step 2. Proving the statement

Consider a positive number M and $N=M^{\left(\frac{1}{k}\right)}$, 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 $N=M^{\left(\frac{1}{k}\right)}$ and simplify,

$$\begin{gathered} x>N \\ =M^{\left(\frac{1}{k}\right)} \end{gathered}$$

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

$$\begin{gathered} x^k>{\left(M^{\left(\frac{1}{k}\right)}\right)}^k \\ {x}^k>M^{\frac{1}{k}·k} \\ {x}^k>M \end{gathered}$$

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

Hence Proved.