Chapter 13.4, Problem 7E

To determine

To find : The limit $\underset{x\to \infty }{\lim }\frac{2x+1}{5x-1}$.

Answer to Problem 7E

  $\underset{x\to \infty }{\lim }\frac{2x+1}{5x-1}=\frac{2}{5}$

Explanation of Solution

Given information:

The given limit is $\underset{x\to \infty }{\lim }\frac{2x+1}{5x-1}$

Calculation :

Let $f$ be a function defined on some interval $\left(a,\infty \right)$ . Then

  $\underset{x\to \infty }{\lim }f\left(x\right)=L$

means that the value of $f\left(x\right)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently large.

Therefore,

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{2x+1}{5x-1}=\underset{x\to \infty }{\lim }\frac{2+\frac{1}{x}}{5-\frac{1}{x}}\text{ (Divide the numerator and denominator by }x\text{)}\\ \underset{x\to \infty }{\lim }\frac{2x+1}{5x-1}=\frac{2+\frac{1}{\infty }}{5-\frac{1}{\infty }}\\ \underset{x\to \infty }{\lim }\frac{2x+1}{5x-1}=\frac{2+0}{5-0}\text{ (Since }\frac{1}{\infty }=\text{0)}\\ \underset{x\to \infty }{\lim }\frac{2x+1}{5x-1}=\frac{2}{5}\end{array}$

Hence,

  $\underset{x\to \infty }{\lim }\frac{2x+1}{5x-1}=\frac{2}{5}$